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Algebra & Equations — Grade 12

Paper 1 revision: identify the method, show the working, check the answer. Build speed with quadratics, inequalities, simultaneous equations and roots.

1. Guided Course 2. Mastery Bank 3. Test Your Knowledge
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Grade 12 CAPS Mathematics

Algebra & Equations

Identify the method • show the maths • check the answer.

The 3-Year Algebra Journey

Grade 12 rewards retrieval, method choice and clean presentation.

10
BuildFactorise • solve • work with expressions
11
ExtendSurds • quadratic methods • inequalities
12
ApplyRecognise • select • verify under exam pressure
Teacher cue
Read the command word first: solve, show, determine, or state. It tells you how much maths to write.

Your Complete Grade 12 Algebra Route

Worth 25±3 of 150 Paper 1 marks — scoring these should feel automatic.

The actual Grade 12 skill
Recognition, under pressure. Grade 11 taught each method one at a time, clearly labelled. Grade 12 just gives an equation with no label — recognising which method it needs, fast, is what this page trains, with many examples per method so even a shaky start still picks up marks. Feels unfamiliar rather than rusty? Go back to Grade 11 Algebra & Equations first.
Complete CAPS route
Also complete Polynomials: Remainder and Factor Theorems and Functions: logarithms for full Grade 12 Algebra coverage.

Diagnostic: What Grade 12 Assumes

If a row here feels unfamiliar, that's exactly what to revise first.

Law / ruleStatementExample
Multiplying powers\(a^m\cdot a^n=a^{m+n}\)\(2^3\cdot2^4=2^7\)
Dividing powers\(a^m\div a^n=a^{m-n}\)\(\dfrac{5^6}{5^2}=5^4\)
Power of a power\((a^m)^n=a^{mn}\)\((3^2)^4=3^8\)
Zero / negative exponent\(a^0=1,\ a^{-n}=\dfrac1{a^n}\)\(7^0=1,\ 2^{-3}=\dfrac18\)
Rational exponent\(a^{\frac{m}{n}}=(\sqrt[n]a)^m\)\(8^{\frac23}=(\sqrt[3]8)^2=4\)
Simplifying a surd\(\sqrt{a\cdot b}=\sqrt a\cdot\sqrt b\)\(\sqrt{50}=5\sqrt2\)
Worked revision — exponential equation Grade 11

Solve for \(x\): \(3^{x-1}=27\)

Show solution
  1. 1Write both sides with the same base: \(27=3^3\), so \(3^{x-1}=3^3\)
  2. 2Equate exponents: \(x-1=3\)
  3. 3\(\boxed{x=4}\) — check: \(3^{4-1}=3^3=27\)✓
Worked revision Grade 11

Solve for \(x\): \(2^{2x-1}=32\)

Show solution
  1. 1Write \(32\) as a power of \(2\): \(32=2^5\)
  2. 2\(2^{2x-1}=2^5\) — same base, so equate exponents: \(2x-1=5\)
  3. 3\(\boxed{x=3}\) — check: \(2^{2(3)-1}=2^5=32\)✓
Common mistake
Writing \(27=3\times9\) and stopping there instead of a single power of \(3\). Every base on both sides must be reduced to the same prime base before exponents can be equated.
Worked revision — surd equation Grade 11

Solve for \(x\): \(\sqrt{2x+3}=x\)

Show solution
  1. 1Square both sides: \(2x+3=x^2\)
  2. 2Rearrange to standard form: \(x^2-2x-3=0\)
  3. 3Factorise: \((x-3)(x+1)=0\Rightarrow x=3\) or \(x=-1\)
  4. 4Check both in the original equation: \(x=3\Rightarrow\sqrt9=3=x\)✓ valid; \(x=-1\Rightarrow\sqrt1=1\neq-1\)✗ rejected
  5. 5\(\boxed{x=3}\) only
Worked revision Grade 11

Solve for \(x\): \(\sqrt{x-1}=x-3\)

Show solution
  1. 1Square both sides: \(x-1=(x-3)^2=x^2-6x+9\)
  2. 2Rearrange to standard form: \(x^2-7x+10=0\)
  3. 3Factorise: \((x-5)(x-2)=0\Rightarrow x=5\) or \(x=2\)
  4. 4Check both in the original equation: \(x=5\Rightarrow\sqrt4=2=5-3\)✓ valid; \(x=2\Rightarrow\sqrt1=1\neq2-3=-1\)✗ rejected
  5. 5\(\boxed{x=5}\) only
Common mistake
Stopping after factorising and giving both roots. Squaring both sides is not reversible — it can create an extraneous root. Every surd equation must be checked back in the original form.
Quick Check

Solve for x: \(2^{x+2}=64\)

Why must every root of a surd equation be checked in the original (unsquared) equation?

The Method-Recognition Routine

This is the actual skill Paper 1 tests. Run every question through these steps before committing to a method.

  1. 1

    What is the variable doing? An exponent (→ exponential equation), under a root (→ surd equation), squared (→ quadratic), in two equations at once (→ simultaneous), or appearing both squared and unsquared, like \(2^{2x}\) alongside \(2^x\) (→ substitution, Method 6)? The variable's position tells you the method family before any algebra.

  2. 2

    Can I get it into a standard, recognisable form? Rearrange to \(ax^2+bx+c=0\), same-base exponents, or "surd alone on one side."

  3. 3

    Which method does this standard form call for? Factorises cleanly → factorise. Ugly numbers or "correct to two decimal places" → quadratic formula. Two unknowns, two equations → substitution.

  4. 4

    Solve, showing the method-mark step. Markers award a mark for the correct method (factors shown, or the formula with values substituted) independently of the final answer.

  5. 5

    Does every root actually work? Surds and variable-in-denominator equations can produce extraneous roots — check in the original equation.

  6. 6

    Is the answer in the form asked for? "Correct to two decimal places," "as a simplified surd," "in the form \(x>a\)" — the mark needs the right format, not just the right number.

Method 1 — Factorisation

Fastest when it works — always try this first.

Worked example Recognise: friendly integers

Solve for \(x\): \(x^2-5x+6=0\)

Show solution
  1. 1Find two numbers that multiply to \(+6\) and add to \(-5\): these are \(-2\) and \(-3\)
  2. 2Write as two factors: \((x-2)(x-3)=0\)
  3. 3Zero-product law: \(x-2=0\) or \(x-3=0\)
  4. 4\(\boxed{x=2\text{ or }x=3}\) — check: \((2)^2-5(2)+6=0\)✓, \((3)^2-5(3)+6=0\)✓
Worked example

Solve for \(x\): \(x^2-3x-10=0\)

Show solution
  1. 1Find two numbers that multiply to \(-10\) and add to \(-3\): these are \(-5\) and \(2\)
  2. 2Write as two factors: \((x-5)(x+2)=0\)
  3. 3Zero-product law: \(x-5=0\) or \(x+2=0\)
  4. 4\(\boxed{x=5\text{ or }x=-2}\) — check: \((5)^2-3(5)-10=0\)✓, \((-2)^2-3(-2)-10=0\)✓
Common mistake
Dividing both sides by \(x\) to "simplify" — this silently deletes the \(x=0\) root whenever it exists. Always move everything to one side and factorise; never divide an equation by the variable.
Recognise it when...
The quadratic has integer coefficients and the numbers "look friendly" — try factorising first, since it's faster than the formula whenever it works.
Your turn — Quick Retrieval (no calculator)

Solve for \(x\): \(x^2+2x-15=0\)

Answer: numbers that multiply to \(-15\) and add to \(2\): \(5\) and \(-3\). \((x+5)(x-3)=0\), so \(x=-5\) or \(x=3\).

Method 1 — Two More Recognisable Shapes

Two traps examiners reuse every year — a hidden common factor, and a negative leading coefficient.

Worked example — common factor first Recognise: a number divides every term

Solve for \(x\): \(2x^2-8x+6=0\)

Show solution
  1. 1Every term shares a factor of \(2\) — take it out first, before trying to factorise the trinomial directly: \(2(x^2-4x+3)=0\)
  2. 2Numbers multiplying to \(3\) and adding to \(-4\): \(-1\) and \(-3\): \(2(x-1)(x-3)=0\)
  3. 3The factor \(2\) can never be zero, so only the brackets matter: \(x-1=0\) or \(x-3=0\)
  4. 4\(\boxed{x=1\text{ or }x=3}\) — check: \(2(1)^2-8(1)+6=0\)✓, \(2(3)^2-8(3)+6=0\)✓
Worked example — negative leading coefficient Recognise: \(-x^2\) leads

Solve for \(x\): \(-x^2+5x-6=0\)

Show solution
  1. 1A negative leading coefficient is awkward to factorise directly — multiply every term by \(-1\) first (this does not change the solutions, since it is the same as multiplying both sides by \(-1\)): \(x^2-5x+6=0\)
  2. 2Numbers multiplying to \(6\) and adding to \(-5\): \(-2\) and \(-3\): \((x-2)(x-3)=0\)
  3. 3Zero-product law: \(x-2=0\) or \(x-3=0\)
  4. 4\(\boxed{x=2\text{ or }x=3}\) — check in the ORIGINAL equation: \(-(2)^2+5(2)-6=0\)✓, \(-(3)^2+5(3)-6=0\)✓
Why multiplying by −1 is always safe here
An equation stays true if both sides are multiplied by the same nonzero number — \(0\times(-1)=0\) too, so the right-hand side is unaffected. This is different from an inequality, where multiplying by a negative number flips the sign; an equation has no direction to flip.

Method 2 — The Quadratic Formula

Always works — use it when factorising doesn't.

Worked example

Solve for \(x\), correct to two decimal places where necessary: \(2x^2+3x-5=0\)

Show solution
  1. 1Identify \(a=2,\ b=3,\ c=-5\)
  2. 2Substitute into \(x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\): \(x=\dfrac{-3\pm\sqrt{9-4(2)(-5)}}{2(2)}=\dfrac{-3\pm\sqrt{49}}{4}\)
  3. 3\(\sqrt{49}=7\): \(x=\dfrac{-3+7}{4}=1\) or \(x=\dfrac{-3-7}{4}=-2{,}5\)
  4. 4\(\boxed{x=1\text{ or }x=-2{,}5}\) — check: \(2(1)^2+3(1)-5=0\)✓
Worked example — irrational roots

Solve for \(x\), correct to two decimal places: \(x^2-4x-3=0\)

Show solution
  1. 1Identify \(a=1,\ b=-4,\ c=-3\)
  2. 2\(\Delta=(-4)^2-4(1)(-3)=16+12=28\) — not a perfect square, so the roots are irrational
  3. 3\(x=\dfrac{4\pm\sqrt{28}}{2}=\dfrac{4\pm2\sqrt7}{2}=2\pm\sqrt7\)
  4. 4\(\boxed{x\approx4{,}65\text{ or }x\approx-0{,}65}\)
Common mistake
Substituting \(-b\) with the wrong sign when \(b\) is already negative. Write \(a\), \(b\), \(c\) as their own line before substituting into the formula.
Recognise it when...
The quadratic doesn't factorise with integers, or the question explicitly asks for an answer "correct to two decimal places" — that instruction signals irrational roots.
Your turn — Quick Retrieval (no calculator)

Write down the discriminant for \(x^2-4x-1=0\), then find the roots.

Answer: \(a=1,b=-4,c=-1\). \(\Delta=16-4(1)(-1)=20\). \(x=\dfrac{4\pm\sqrt{20}}{2}=2\pm\sqrt5\).

Method 2 — Two More Recognisable Shapes

An equation that is not yet in standard form, and a case where the formula reveals there is no real answer at all.

Worked example — rearrange before you identify a, b, c Recognise: variable on both sides

Solve for \(x\): \(3x^2=2-5x\)

Show solution
  1. 1The equation is not in standard form yet — move every term to one side before reading off \(a\), \(b\), \(c\): \(3x^2+5x-2=0\)
  2. 2Identify \(a=3,\ b=5,\ c=-2\)
  3. 3Substitute: \(x=\dfrac{-5\pm\sqrt{25-4(3)(-2)}}{2(3)}=\dfrac{-5\pm\sqrt{49}}{6}=\dfrac{-5\pm7}{6}\)
  4. 4\(\boxed{x=\dfrac13\text{ or }x=-2}\) — check: \(3\left(\dfrac13\right)^2+5\left(\dfrac13\right)-2=0\)✓
Worked example — no real solutions Recognise: the formula itself gives the answer

Solve for \(x\): \(x^2+2x+5=0\)

Show solution
  1. 1Identify \(a=1,\ b=2,\ c=5\) and compute the discriminant before substituting the rest of the formula — it decides whether a real answer even exists
  2. 2\(\Delta=(2)^2-4(1)(5)=4-20=-16\)
  3. 3\(\Delta<0\) means the formula is asking for the square root of a negative number, which has no real value
  4. 4\(\boxed{\text{no real solutions}}\) — do not write \(x=\pm\sqrt{-16}\) or leave the answer half-finished; state the conclusion in words
Recognise it when...
The equation is not written as "\(\ldots=0\)" yet — always rearrange to standard form before reading off \(a\), \(b\), \(c\), never partway through substituting. And if \(\Delta<0\) turns up, that is a complete, correct answer on its own — there is nothing left to compute.
Quick Check

Which method should you try FIRST for \(x^2-7x+12=0\)?

Solve for x, correct to two decimal places: \(x^2-3x-1=0\)

Solve for x: \(3x^2=2-5x\)

Method 3 — Quadratic Inequalities

Solve the equation for critical values, then read the sign from the parabola's shape.

Worked example

Solve for \(x\): \(x^2-x-6\le0\)

Show solution
  1. 1Factorise as if solving the equation: \(x^2-x-6=0\Rightarrow(x-3)(x+2)=0\Rightarrow x=3\) or \(x=-2\) (critical values)
  2. 2\(a=1>0\), so the parabola opens upward — it is \(\le0\) between its roots
  3. 3\(\boxed{-2\le x\le3}\) (closed circles, since \(\le\), not \(<\))
  4. 4Check with \(x=0\) (inside the interval): \((0)^2-0-6=-6\le0\)✓
A number line showing the solution -2 to 3 inclusiveA number line from -4 to 5 with closed circles at -2 and 3, and the segment between them highlighted, showing the inequality is satisfied between the two roots.−4−2035
Closed circles at \(-2\) and \(3\) show the endpoints are included (\(\le\)), matching the solid segment joining them.
Worked example — the other direction

Solve for \(x\): \(x^2-5x+4>0\)

Show solution
  1. 1Solve the associated equation: \(x^2-5x+4=0\Rightarrow(x-1)(x-4)=0\Rightarrow x=1\) or \(x=4\)
  2. 2\(a=1>0\), so the parabola opens upward — it is \(>0\) outside its roots this time, the opposite region from the first example above
  3. 3\(\boxed{x<1\text{ or }x>4}\)
Common mistake
Writing "\(x\le-2\) or \(x\le3\)" by treating it like a linear inequality. A quadratic inequality's solution is a region relative to the parabola's shape, not two separate one-sided inequalities — reason about the direction it opens first.
Your turn — Quick Retrieval (no calculator)

Solve for \(x\): \(x^2-9>0\)

Answer: \((x-3)(x+3)=0\Rightarrow x=\pm3\). Upward parabola, \(>0\) outside the roots: \(x<-3\) or \(x>3\).

Method 3 — Two More Recognisable Shapes

An inequality that needs rearranging first, and the repeated-root edge case that breaks the usual pattern.

Worked example — rearrange first Recognise: terms on both sides

Solve for \(x\): \(2x^2\ge5x+3\)

Show solution
  1. 1Move every term to one side so it compares to zero — the shape of the parabola only means something once one side is \(0\): \(2x^2-5x-3\ge0\)
  2. 2Solve the associated equation to find the critical values: \(2\times(-3)=-6\); numbers multiplying to \(-6\) and adding to \(-5\) are \(-6\) and \(1\): \(2x^2-6x+x-3=2x(x-3)+1(x-3)=(2x+1)(x-3)=0\Rightarrow x=-\dfrac12\) or \(x=3\)
  3. 3\(a=2>0\), so the parabola opens upward — it is \(\ge0\) outside its roots
  4. 4\(\boxed{x\le-\dfrac12\text{ or }x\ge3}\) — check with \(x=0\) (should fail, since \(0\) sits between the roots): \(2(0)=0\), \(5(0)+3=3\), and \(0\ge3\) is false✓, confirming \(0\) is correctly excluded
Worked example — the repeated-root edge case Recognise: a perfect square trinomial

Solve for \(x\): \(x^2-6x+9\le0\)

Show solution
  1. 1Recognise this as a perfect square: \(x^2-6x+9=(x-3)^2\), so the inequality is \((x-3)^2\le0\)
  2. 2A squared real number is never negative — the smallest value \((x-3)^2\) can ever take is \(0\), exactly at \(x=3\)
  3. 3So "\(\le0\)" is only possible at the single point where equality holds — there is no region either side of the root this time
  4. 4\(\boxed{x=3}\) only — check: at \(x=4\), \((4-3)^2=1\), and \(1\le0\) is false, confirming no other value works
Recognise it when...
The two numbers you find for the critical values turn out identical — that is the discriminant-zero case in disguise. A repeated root only ever gives a single-point solution (for \(\le\) or \(\ge\)) or no solution at all (for \(<\) or \(>\)), never the usual "between" or "outside" region.

Method 4 — Simultaneous Equations

One linear, one quadratic — substitute the linear equation into the quadratic one.

Worked example

Solve simultaneously for \(x\) and \(y\): \(y=x+2\) and \(y=x^2\)

Show solution
  1. 1Both expressions equal \(y\), so set them equal: \(x+2=x^2\)
  2. 2Rearrange to standard form: \(x^2-x-2=0\)
  3. 3Factorise: \((x-2)(x+1)=0\Rightarrow x=2\) or \(x=-1\)
  4. 4Substitute each \(x\) into the linear equation (simplest): \(x=2\Rightarrow y=4\); \(x=-1\Rightarrow y=1\)
  5. 5\(\boxed{(2;4)\text{ and }(-1;1)}\) — check both in \(y=x^2\): \(4=(2)^2\)✓, \(1=(-1)^2\)✓
Worked example — rearrange first

Solve simultaneously for \(x\) and \(y\): \(x-y=1\) and \(y=x^2-1\)

Show solution
  1. 1The linear equation isn't in \(y=\ldots\) form yet — rearrange it first: \(x-y=1\Rightarrow y=x-1\)
  2. 2Substitute into the quadratic equation: \(x-1=x^2-1\)
  3. 3Rearrange to standard form: \(0=x^2-x\)
  4. 4Factorise: \(x(x-1)=0\Rightarrow x=0\) or \(x=1\)
  5. 5Substitute each \(x\) into \(y=x-1\): \(x=0\Rightarrow y=-1\); \(x=1\Rightarrow y=0\)
  6. 6\(\boxed{(0;-1)\text{ and }(1;0)}\)
A line and a parabola intersecting at two pointsThe line y equals x plus 2 and the parabola y equals x squared, intersecting at (-1,1) and (2,4).(−1,1)(2,4)y=x²y=x+2
The two intersection points are exactly the two solutions found algebraically — a line and a parabola can cross at most twice.
Common mistake
Finding \(x\) and stopping. A simultaneous-equations question always wants the matching \(y\)-value too — give every \((x;y)\) pair, not just the \(x\)-values.
Your turn — Quick Retrieval (no calculator)

Solve simultaneously: \(y=2x-1\) and \(y=x^2-4\)

Answer: \(2x-1=x^2-4\Rightarrow x^2-2x-3=0\Rightarrow(x-3)(x+1)=0\Rightarrow x=3\) or \(x=-1\). Then \(y=5\) or \(y=-3\).

Method 4 — Woven Into a Word Problem

Exactly the same two steps — the only new skill is translating the words into the two equations first.

Worked example — a real-world context Recognise: two unknowns, two conditions

A rectangular garden has a perimeter of \(20\) m and an area of \(24\text{ m}^2\). Determine its length and width.

Show solution
  1. 1Let the length be \(x\) m and the width be \(y\) m. "Perimeter \(20\)" gives \(2x+2y=20\), i.e. \(x+y=10\); "area \(24\)" gives \(xy=24\) — two unknowns, two equations, exactly the Method 4 shape once translated
  2. 2Make \(y\) the subject of the simpler equation: \(y=10-x\)
  3. 3Substitute into the other equation and rearrange to standard form: \(x(10-x)=24\Rightarrow10x-x^2=24\Rightarrow x^2-10x+24=0\)
  4. 4Factorise: \((x-4)(x-6)=0\Rightarrow x=4\) or \(x=6\)
  5. 5Both solutions describe the same rectangle from opposite sides — if \(x=4\) then \(y=6\), and if \(x=6\) then \(y=4\): \(\boxed{\text{the rectangle is }4\text{ m by }6\text{ m}}\) (check: \(2(4)+2(6)=20\)✓, \(4\times6=24\)✓)
Worked example — a second linear-quadratic pair

Solve simultaneously for \(x\) and \(y\): \(y=3x-2\) and \(y=x^2-x-2\)

Show solution
  1. 1Both expressions equal \(y\), so set them equal: \(3x-2=x^2-x-2\)
  2. 2Rearrange to standard form: \(0=x^2-4x\)
  3. 3Factorise: \(x(x-4)=0\Rightarrow x=0\) or \(x=4\)
  4. 4Substitute each \(x\) into the linear equation: \(x=0\Rightarrow y=-2\); \(x=4\Rightarrow y=10\)
  5. 5\(\boxed{(0;-2)\text{ and }(4;10)}\) — check both in \(y=x^2-x-2\): \(-2=0-0-2\)✓, \(10=16-4-2\)✓
Recognise it when...
A word problem hands you exactly two unknown quantities and exactly two conditions relating them — that is the Method 4 shape wearing a real-world disguise. Translate each condition into an equation first, then it is identical algebra to every other example above.
Quick Check

Solve for x: \(x^2+2x-8\ge0\)

When solving one linear and one quadratic equation simultaneously, why substitute into the LINEAR equation to find y?

Solve for x: \(2x^2\ge5x+3\)

Method 5 — Nature of Roots

The discriminant only — no solving needed.

DiscriminantNature of roots
\(\Delta>0\), perfect squareReal, unequal, rational
\(\Delta>0\), not a perfect squareReal, unequal, irrational
\(\Delta=0\)Real and equal
\(\Delta<0\)No real roots
Δ<0 — no real roots

\(x^2+x+1=0\)

\(\Delta=-3\)

Δ=0 — equal roots

\(x^2-6x+9=0\)

x=3

\(\Delta=0\)

Δ>0 — two real roots

\(x^2-3x-4=0\)

x=-1x=4

\(\Delta=25\) (perfect square, so rational)

Why this works
The discriminant is exactly the part under the square root in the quadratic formula. If it's negative, the formula asks for the square root of a negative number — impossible for real numbers, so the parabola never reaches the x-axis. If it's zero, the “±” in the formula collapses to nothing — the two roots merge into one, exactly where the vertex touches the axis. If it's positive, the formula genuinely gives two different values — the parabola crosses the axis twice.
Worked example A real CAPS clarification example

Show that \(x^2+x+1=0\) has no real roots.

Show solution
  1. 1You don't need to solve the equation — only evaluate \(\Delta=b^2-4ac\)
  2. 2\(a=1,b=1,c=1\): \(\Delta=(1)^2-4(1)(1)=1-4=-3\)
  3. 3\(\Delta<0\), so \(\boxed{\text{there are no real roots}}\) (a negative number has no real square root)
Worked example — equal roots

Determine the nature of the roots of \(x^2-6x+9=0\), without solving the equation.

Show solution
  1. 1\(a=1,b=-6,c=9\): \(\Delta=(-6)^2-4(1)(9)=36-36=0\)
  2. 2\(\Delta=0\), so \(\boxed{\text{the roots are real and equal}}\) (this expression is in fact a perfect square: \(x^2-6x+9=(x-3)^2\))
Worked example — two distinct real roots

Determine the nature of the roots of \(x^2-3x-4=0\), without solving the equation.

Show solution
  1. 1\(a=1,b=-3,c=-4\): \(\Delta=(-3)^2-4(1)(-4)=9+16=25\)
  2. 2\(\Delta>0\), so there are two distinct real roots; \(25\) is a perfect square, so \(\boxed{\text{the roots are also rational}}\) (it factorises: \((x-4)(x+1)=0\))
Common mistake
Solving the whole equation with the quadratic formula when only asked about the nature of the roots. Calculating \(\Delta\) and stating its sign is the entire method.
Recognise it when...
The question says "show that," "determine the nature of," or "for which value(s) of \(k\) will the equation have equal/real/non-real roots" — all discriminant questions, not solving questions.
Your turn — Quick Retrieval (no calculator)

Show that the roots of \(x^2-2x-1=0\) are real, irrational and unequal.

Answer: \(\Delta=(-2)^2-4(1)(-1)=4+4=8\). \(\Delta>0\) so real and unequal; \(8\) is not a perfect square, so irrational.

Method 5 — Two More Recognisable Shapes

A "prove real for every value" question, and a "find the values of k" question that turns into its own quadratic inequality.

Worked example — prove real for every value Recognise: "prove" or "show that ... for all"

Prove that \(x^2-2kx+(k^2-1)=0\) has real roots for all real values of \(k\).

Show solution
  1. 1"Real roots" for every value of a parameter is always a discriminant question — compute \(\Delta\) in terms of \(k\): \(a=1,\ b=-2k,\ c=k^2-1\)
  2. 2\(\Delta=(-2k)^2-4(1)(k^2-1)=4k^2-4k^2+4\)
  3. 3Every \(k^2\) term cancels, leaving a constant that does not depend on \(k\) at all: \(\boxed{\Delta=4}\)
  4. 4Since \(\Delta=4>0\) regardless of which value \(k\) takes, the equation has two distinct real roots for every real \(k\) — this is exactly what "prove real for all values of \(k\)" is asking for
Worked example — nature of roots leading to an inequality Recognise: "find the values of k"

Determine the value(s) of \(k\) for which \(x^2+(k-2)x+(k+1)=0\) has non-real roots.

Show solution
  1. 1Non-real roots means \(\Delta<0\) — set up the discriminant in terms of \(k\): \(a=1,\ b=(k-2),\ c=(k+1)\)
  2. 2\(\Delta=(k-2)^2-4(1)(k+1)=k^2-4k+4-4k-4=k^2-8k\)
  3. 3The condition itself is now a quadratic inequality in \(k\), solved exactly like Method 3: \(k^2-8k<0\Rightarrow k(k-8)<0\)
  4. 4Critical values \(k=0\) and \(k=8\); this "parabola in \(k\)" opens upward, so it is negative between its roots
  5. 5\(\boxed{0
Why this is worth extra attention
A "for which values of \(k\)" nature-of-roots question is really two methods stacked together: set up \(\Delta\) using Method 5's own formula, then solve the resulting inequality in \(k\) exactly like Method 3. Recognising that stack — not either method alone — is what separates a full-marks answer from a half-finished one.
Quick Check

For \(2x^2-4x+2=0\), what is the nature of the roots?

A question asks: "for which value(s) of k will \(x^2+kx+9=0\) have equal roots?" What should you compute?

For \(x^2-2kx+(k^2-1)=0\), which statement is correct?

Method 6 — Equations Reducible to Quadratic Form

Not always a quadratic on the surface — but one substitution turns it into a familiar Method 1 or 2 equation.

Not a separate CAPS knowledge statement
This is not its own listed CAPS skill — it is exponent equations (Grade 10/11) or a higher-degree equation with only even powers, combined directly with quadratic-solving (Grade 11). NSC papers reuse this combination often, and it is exactly the kind of unlabelled combination the whole page trains you to spot.
The tell

Look for a repeated "building block" — the same expression appearing squared and unsquared, e.g. \(2^{2x}\) is \((2^x)^2\), or \(x^4\) is \((x^2)^2\). Let a new letter stand for that block, solve the resulting quadratic in the new letter, then convert back.

Method
  • Spot the repeated block and let \(y=\) that block.
  • Rewrite the whole equation in terms of \(y\) — it should now look exactly like a standard quadratic.
  • Solve for \(y\) using Method 1 or 2 above.
  • Convert every valid \(y\)-value back to \(x\), rejecting any \(y\) the block could never actually equal.
Worked example — exponential-quadratic Recognise: the exponent is doubled somewhere

Solve for \(x\): \(2^{2x}-6(2^x)+8=0\)

Show solution
  1. 1Recognise the repeated block: \(2^{2x}=(2^x)^2\), so the whole equation is built from \(2^x\) appearing squared and unsquared
  2. 2Let \(y=2^x\). Rewriting the equation: \(y^2-6y+8=0\)
  3. 3Factorise as a normal quadratic: \((y-2)(y-4)=0\Rightarrow y=2\) or \(y=4\)
  4. 4Convert back: since \(2^x\) is always positive, both values are valid candidates. \(2^x=2\Rightarrow x=1\). \(2^x=4\Rightarrow x=2\)
  5. 5\(\boxed{x=1\text{ or }x=2}\) — check: \(2^{2(1)}-6(2^1)+8=4-12+8=0\)✓, \(2^{2(2)}-6(2^2)+8=16-24+8=0\)✓
Worked example — a quartic in disguise Recognise: only even powers of x

Solve for \(x\): \(x^4-13x^2+36=0\)

Show solution
  1. 1Only even powers of \(x\) appear, and \(x^4=(x^2)^2\) — the repeated block is \(x^2\)
  2. 2Let \(y=x^2\). Rewriting the equation: \(y^2-13y+36=0\)
  3. 3Numbers multiplying to \(36\) and adding to \(-13\): \(-4\) and \(-9\): \((y-4)(y-9)=0\Rightarrow y=4\) or \(y=9\)
  4. 4Convert back: \(x^2=4\Rightarrow x=\pm2\). \(x^2=9\Rightarrow x=\pm3\) — both are valid, since \(x^2\) can equal any positive number
  5. 5\(\boxed{x=\pm2\text{ or }x=\pm3}\) — four solutions in total, check one: \((2)^4-13(2)^2+36=16-52+36=0\)✓
Common mistake
Stopping once \(y\) is found, and forgetting to convert back to \(x\). The question asked to solve for \(x\), not \(y\) — \(y\) is only ever a stepping stone.
Recognise it when...
The equation is not linear in its "obvious" variable, but one expression clearly appears both squared and unsquared — an exponent that is double another exponent in the same equation, or a degree-4 equation with no odd-power terms at all.
Your turn — Quick Retrieval (no calculator)

Solve for \(x\): \(3^{2x}-4(3^x)+3=0\)

Answer: let \(y=3^x\): \(y^2-4y+3=0\Rightarrow(y-1)(y-3)=0\Rightarrow y=1\) or \(y=3\). \(3^x=1\Rightarrow x=0\); \(3^x=3\Rightarrow x=1\).

Quick Check

To solve \(2^{2x}-6(2^x)+8=0\), which substitution turns it into a quadratic?

Solve for x: \(x^4-13x^2+36=0\)

Integrated NSC Rider

A full Paper 1 question combines these methods with no labels — this is what recognition looks like under real conditions.

Full rider 7 marks

Given: \(y=x-1\) and \(x^2+xy=6\).

(1) Solve for \(x\) and \(y\) simultaneously. (5)

(2) Hence, state for which value(s) of \(x\) the product \(xy\) is negative. (2)

Show solution
  1. 1Substitute \(y=x-1\) into the second equation: \(x^2+x(x-1)=6\Rightarrow x^2+x^2-x=6\Rightarrow2x^2-x-6=0\)
  2. 2Factorise: \((2x+3)(x-2)=0\Rightarrow x=2\) or \(x=-\dfrac32\)
  3. 3Substitute back into \(y=x-1\): \(x=2\Rightarrow y=1\); \(x=-\dfrac32\Rightarrow y=-\dfrac52\)
  4. 4\(\boxed{(2;1)\text{ and }(-\tfrac32;-\tfrac52)}\)
  5. 5For part (2): \(xy=2(1)=2>0\) for the first pair; \(xy=(-\tfrac32)(-\tfrac52)=\tfrac{15}4>0\) for the second — neither solution gives a negative product
  6. 6\(\boxed{\text{no value of }x\text{ among the solutions gives }xy<0}\)
Reading the rider
Notice this question never says "simultaneous equations" or "quadratic" — the routine (variable position, standard form, method match) is what tells you \(y=x-1\) substitutes into a quadratic, exactly like Method 4 above. Part (2) then reuses the actual coordinate pairs you already found, no new algebra required.
Full rider — your turn 6 marks

Given: \(y=x+2\) and \(xy=8\).

(1) Solve for \(x\) and \(y\) simultaneously. (4)

(2) Hence, state which solution has both \(x\) and \(y\) positive. (2)

Show solution
  1. 1Substitute \(y=x+2\) into \(xy=8\): \(x(x+2)=8\Rightarrow x^2+2x-8=0\)
  2. 2Factorise: \((x+4)(x-2)=0\Rightarrow x=-4\) or \(x=2\)
  3. 3Substitute back into \(y=x+2\): \(x=-4\Rightarrow y=-2\); \(x=2\Rightarrow y=4\)
  4. 4\(\boxed{(-4;-2)\text{ and }(2;4)}\)
  5. 5For part (2): only \((2;4)\) has both coordinates positive — \((-4;-2)\) has both negative
  6. 6\(\boxed{x=2,\ y=4}\)

Integrated NSC Rider — Two More

A system built from two quadratics, and a two-part question that chains Method 5 straight into Method 1.

Full rider 6 marks

Given: \(x+y=7\) and \(x^2+y^2=25\).

Solve for \(x\) and \(y\) simultaneously.

Show solution
  1. 1Both equations involve \(x\) and \(y\) with no obvious label — the variable position (one equation linear, the other has squared terms) is what flags this as Method 4, even though neither equation is written as "\(y=\ldots\)" the way earlier examples were
  2. 2Make \(y\) the subject of the linear equation: \(y=7-x\)
  3. 3Substitute into the second equation and expand carefully: \(x^2+(7-x)^2=25\Rightarrow x^2+49-14x+x^2=25\)
  4. 4Collect like terms and rearrange to standard form: \(2x^2-14x+24=0\Rightarrow x^2-7x+12=0\)
  5. 5Factorise: \((x-3)(x-4)=0\Rightarrow x=3\) or \(x=4\)
  6. 6Substitute back into \(y=7-x\): \(x=3\Rightarrow y=4\); \(x=4\Rightarrow y=3\)
  7. 7\(\boxed{(3;4)\text{ and }(4;3)}\) — check: \(3+4=7\)✓, \(3^2+4^2=9+16=25\)✓
Reading the rider
CAPS describes this method family as "one linear, one quadratic" — \(x^2+y^2=25\) is genuinely quadratic (in two variables), it just is not written as \(y=x^2+\ldots\) the way Method 4's own worked examples were. The substitution step is identical either way; only the amount of expanding after substituting changes.
Full rider — your turn 5 marks

Given: \(x^2+px+4=0\).

(1) Determine the value(s) of \(p\) for which the equation has equal roots. (3)

(2) Hence, solve for \(x\) for each value of \(p\) found in (1). (2)

Show solution
  1. 1"Equal roots" is a Method 5 question first: \(a=1,\ b=p,\ c=4\), so \(\Delta=p^2-4(1)(4)=p^2-16\)
  2. 2Equal roots means \(\Delta=0\): \(p^2-16=0\Rightarrow p^2=16\Rightarrow\boxed{p=4\text{ or }p=-4}\)
  3. 3"Hence" means reuse this result — substitute each \(p\)-value back into the original equation and solve with Method 1, since equal roots always factorise as a perfect square
  4. 4For \(p=4\): \(x^2+4x+4=0\Rightarrow(x+2)^2=0\Rightarrow x=-2\)
  5. 5For \(p=-4\): \(x^2-4x+4=0\Rightarrow(x-2)^2=0\Rightarrow x=2\)
  6. 6\(\boxed{p=4\Rightarrow x=-2;\quad p=-4\Rightarrow x=2}\)

Grade 12 Method Sprint

Name the method before you solve. That is the exam skill.

01 Solve: \(x^2-7x+10>0\)
Factorise: \((x-5)(x-2)>0\Rightarrow x<2\text{ or }x>5\)
02 Solve: \(2x^2+x-3=0\)
Factorise: \((2x+3)(x-1)=0\Rightarrow x=1\text{ or }-\dfrac32\)
03 Solve: \(y=2x+1\), \(y=x^2\)
Substitute: \(x^2=2x+1\Rightarrow x=1\pm\sqrt2\).
04 Nature: \(4x^2-4x+1=0\)
Discriminant: \(\Delta=(-4)^2-4(4)(1)=0\): real and equal.
05 Solve: \(3^{2x}-4(3^x)+3=0\)
Substitute \(y=3^x\): \((y-1)(y-3)=0\Rightarrow x=0\text{ or }1\)
06 Prove real for all k: \(x^2-2kx+(k^2-1)=0\)
Discriminant: \(\Delta=4k^2-4(k^2-1)=4>0\) for every \(k\)

Exam Strategy

Checklist before you submit
  • Ran the 6-step recognition routine before committing to a method?
  • Tried factorising before reaching for the quadratic formula?
  • Equation not yet "\(\ldots=0\)"? Rearranged to standard form before reading off \(a\), \(b\), \(c\)?
  • Quadratic inequality: reasoned about the parabola's direction, not treated it like a linear inequality?
  • Simultaneous equations: gave every \((x;y)\) pair, not just the x-values?
  • Nature of roots: only computed \(\Delta\), didn't fully solve when not asked to?
  • Equation has the same expression squared and unsquared (e.g. \(2^{2x}\) and \(2^x\), or \(x^4\) and \(x^2\))? Tried a substitution before assuming it is unsolvable?
  • Checked any surd-equation root in the ORIGINAL equation?
Exam mantra
Recognise. Recall. Execute. Check the format. Every method here was taught in full in Grade 11 — the only new skill this year is doing all four steps fast, unlabelled, inside a bigger question.
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What CAPS Actually Expects Here

Not a new knowledge statement — a scope clarification.

CAPS assigns exponents, surds, and solving quadratic equations/inequalities to Grade 11 Term 1 — there is no dedicated Grade 12 Algebra knowledge statement. What Grade 12 Paper 1 actually examines is these Grade 11 skills applied fluently and in combination, most often as the opening question(s) of the paper and folded into Finance, Sequences, Calculus and Trigonometry riders elsewhere in the paper.

  • Quadratic equations by factorisation and by the quadratic formula.
  • Quadratic inequalities, interpreted via the parabola's shape.
  • Simultaneous equations, one linear and one quadratic (in two variables).
  • Nature of roots, using the discriminant, without necessarily solving.
  • Equations reducible to quadratic form via substitution (e.g. exponential-quadratic or quartic-in-\(x^2\) patterns) — not a separate CAPS statement, but a real, commonly examined way Paper 1 combines exponents with quadratic-solving.

How to Use This Lesson

A few practical notes before you start.

  • If the diagnostic table or its two worked examples feel genuinely unfamiliar, use the full Grade 11 Algebra & Equations page first — this page assumes that teaching already happened.
  • Cover the solution and try the "Your turn" retrieval item under each method before revealing the worked example's answer.
  • Run the 6-step Method-Recognition Routine on every practice question, even ones you think you already recognise — building the habit is the point.
  • The Integrated NSC Rider is deliberately unlabelled — that's the real exam experience, not a simplification.

Learn It in Short Videos

Four free, independent videos — not made by Equation Station SA.

The Quadratic Formula

Using the formula when factorising doesn't work.

Khan Academy · Using the quadratic formula

Quadratic Inequalities

Reading the sign from the parabola's shape.

Khan Academy · Quadratic inequalities

Simultaneous Equations

One linear, one quadratic, solved by substitution.

Khan Academy · Non-linear systems of equations

Nature of Roots

Using the discriminant without solving.

Khan Academy · Discriminant for types of solutions

Algebra Study Path

Learn → practise → visualise → stretch.

Step 1 · Learn

Siyavula: Grade 11 Mathematics

Rebuild the Grade 11 foundations first.

Open Siyavula
Step 2 · Practise

Khan Academy: Quadratics

Practise quadratics, systems and inequalities.

Study quadratics
Step 3 · Visualise

GeoGebra Graphing Calculator

Plot roots, turning points and intersections.

Open GeoGebra
Step 4 · Extend

DBE CAPS Learner Books

Use the official Grade 12 learner book for extra CAPS examples.

Open DBE books

Common Exam Mistakes

Avoid these errors. They cost marks every year.

Dividing an equation by the variable

Silently deletes the \(x=0\) root whenever it exists — always factorise instead.

Not checking surd-equation roots

Squaring can introduce an extraneous root — always substitute back into the original equation.

Treating a quadratic inequality like a linear one

The solution is a region relative to the parabola's shape — reason about the direction it opens first.

Giving x-values without matching y-values

A simultaneous-equations answer needs every full \((x;y)\) pair, not just the x-solutions.

Fully solving when only asked for the nature of roots

Computing \(\Delta\) and stating its sign is the entire method — wastes time and risks new errors.

Ignoring the requested answer format

"Correct to two decimal places," "as a simplified surd" — the mark needs the right format, not just the right number.

This Topic in Other Grades

Algebra & Equations builds across all three FET grades — revisit the earlier grades here.

Grade 10

Real numbers, factorising, algebraic fractions, exponent laws, and linear/quadratic/simultaneous equations

Study this

Grade 11

Rational exponents, surds, completing the square, the quadratic formula, quadratic inequalities and nature of roots — taught from scratch

Study this

Practise This Topic

You've done the notes above — now practise and test yourself.

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Grade 12 Algebra & Equations Mastery Bank

Exam-style Grade 12 questions arranged by level, combining original practice with clearly identified paper-and-memo matches.

Step 1 • Practise • Cognitive Levels
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Algebra & Equations Grade 12 Test Your Knowledge

Auto-marked quiz with instant feedback, explanations and a complete answer review.

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Student Questions

Frequently Asked Questions

Straight answers to common Grade 12 CAPS questions about algebra and equations.

Is there new Algebra content in Grade 12?

No. CAPS is explicit that exponents, surds, and solving quadratic equations and inequalities are Grade 11 Term 1 content — there is no separate "Algebra" topic taught for the first time in Grade 12. Instead, these skills are assumed fluent and tested by being woven into Grade 12 finance, sequences, calculus and trigonometry questions, and directly in Paper 1's opening questions.

So why does this page exist if there's nothing new to learn?

Because the actual Grade 12 skill isn't learning a new method — it's recognising, fast and under exam pressure, which of five Grade 11 methods a given question needs, then executing it without the guided context Grade 11 provided. This page is built around that recognition skill, not a first-time re-teach.

What are the five methods this page covers?

Quadratic equations by factorisation, the quadratic formula, quadratic inequalities, simultaneous equations (one linear, one quadratic), and determining the nature of roots using the discriminant.

I'm shaky on the Grade 11 basics — where should I start?

Start with this page's diagnostic table and two worked revision examples. If either feels unfamiliar, go back to the full Grade 11 Algebra & Equations page first — it teaches these skills from scratch, with far more worked examples and practice than a revision page can fit.

How is this different from the Grade 11 Algebra page?

The Grade 11 page teaches each method individually with a full first-time explanation. This Grade 12 page assumes that teaching already happened and instead trains you to recognise which method a mixed, unlabelled question needs, and to execute it inside a full multi-part exam rider.

What mistakes should I avoid?

Dividing an equation by the variable (silently loses a root), forgetting to check surd-equation solutions in the original equation, treating a quadratic inequality like a linear one, forgetting to give the matching y-value in a simultaneous-equations answer, and fully solving an equation when the question only asked for the nature of its roots.