Algebra & Equations — Grade 11

Q: What does the discriminant tell you?

The discriminant, b^2-4ac, tells you the nature of a quadratic equation's roots without solving it: positive means two distinct real roots, zero means one repeated (equal) real root, and negative means no real roots.

Build fluency: choose the method, show every line, check every root. Master exponents, surds, quadratics and inequalities.

1. Guided Course 2. Mastery Bank 3. Test Your Knowledge

CAPS alignedGrade 11Paper 1Interactive notes

Grade 11 CAPS Mathematics

Algebra & Equations

Learn the rule • model the method • practise with feedback.

The 3-Year Algebra Journey

Grade 11 turns Grade 10 tools into quadratic fluency.

BuildFactorise • solve • manipulate expressions

ExtendSurds • quadratics • formula • inequalities

ApplyRecognise the method in a mixed exam question

Teacher cue

Standard form first: write (ax^2+bx+c=0), then choose the method.

The Biggest Topic This Year

Algebra & Equations is worth 45±3 of 150 Paper 1 marks — the single highest-weighted topic in the whole Grade 11 curriculum.

The big idea

This year extends everything from Grade 10: exponents now include fractions, surds become something you can solve equations with, and quadratics get two brand-new solving methods that work even when factorising doesn't.

Where we're headed

Rational exponents → surd arithmetic → surd equations → completing the square → the quadratic formula → quadratic inequalities → simultaneous linear-quadratic equations → nature of roots. Each builds on the one before it.

Rational Exponents

A fractional exponent is a root in disguise.

Quick recall from Grade 10

Integer exponent laws still apply this year, e.g. \((x^m)^n=x^{mn}\). And root notation \(\sqrt[q]{x}\) means "the number that, raised to the power \(q\), gives back \(x\)" — e.g. \(\sqrt[3]{8}=2\) because \(2^3=8\).

Where the rule below comes from

Using the power-of-a-power law above: \(\left(x^{\frac12}\right)^2=x^{\frac12\times2}=x^1=x\). So \(x^{\frac12}\), squared, gives back \(x\) — exactly the definition of a square root. That's why a power of \(\frac1q\) is defined to mean a \(q\)-th root, not an arbitrary new rule to memorise.

The rule

x^{\frac{p}{q}}=\sqrt[q]{x^p}=\left(\sqrt[q]{x}\right)^p\qquad(x>0,\ q>0)

Reading it

The denominator of the fractional exponent is the root (the index), and the numerator is the power. It's usually easier to take the root first, then raise to the power — that keeps the numbers small.

Worked example — simplest case Level 1

Evaluate: \(4^{\frac{1}{2}}\)

Show solution

1The denominator \(2\) is the root, and the numerator \(1\) leaves it unchanged: \(4^{\frac12}=\sqrt4\)
2\(\boxed{2}\)

Worked example Level 2

Evaluate: \(9^{\frac{3}{2}}\)

Show solution

1The denominator \(2\) is the root; take the square root of \(9\) first: \(\sqrt9=3\).
2Raise the result to the numerator's power, \(3\): \(3^3=27\).
3\(\boxed{9^{\frac32}=27}\)

Worked example Level 2

Evaluate: \(8^{\frac{2}{3}}\)

Show solution

1Take the cube root of \(8\) first: \(\sqrt[3]{8}=2\).
2Raise to the power \(2\): \(2^2=4\).
3\(\boxed{8^{\frac23}=4}\)

Worked example — solving an equation Level 3

Solve for \(x\): \(x^{\frac23}=4\)

Show solution

1Raise both sides to the power \(\frac32\) (the reciprocal of \(\frac23\)), so the left side becomes \(x^{\frac23\times\frac32}=x^1=x\): \(x=4^{\frac32}\)
2Take the square root of \(4\) first, then cube the result: \(\sqrt4=2,\ 2^3=8\)
3\(\boxed{x=8}\) — check: \(8^{\frac23}=(\sqrt[3]8)^2=2^2=4\)✓

Quick Check

Evaluate: \(4^{\frac32}\)

Evaluate: \(27^{\frac13}\)

Adding, Subtracting and Multiplying Surds

Simplify first — then only like surds can combine.

What is a surd?

A surd is a root that is irrational — it can never be simplified to a whole number or fraction, like \(\sqrt2\) or \(\sqrt5\). \(\sqrt4\) is not a surd, because \(\sqrt4=2\) simplifies to a whole number.

Multiplying

\sqrt{a}\times\sqrt{b}=\sqrt{ab}

Dividing

\dfrac{\sqrt{a}}{\sqrt{b}}=\sqrt{\dfrac{a}{b}}

Simplify first

Using the multiplying rule in reverse, extract any perfect square factor: \(\sqrt{50}=\sqrt{25\times2}=\sqrt{25}\times\sqrt2=5\sqrt2\).

Adding/subtracting

Only like surds (same number under the root) can be combined, exactly like like-terms: \(3\sqrt2+5\sqrt2=8\sqrt2\).

Worked example — already like surds Level 1

Simplify: \(2\sqrt5+3\sqrt5\)

Show solution

1Both terms already have \(\sqrt5\), so add the coefficients directly, the same way you'd add \(2x+3x\).
2\(\boxed{5\sqrt5}\)

Worked example Level 2

Simplify: \(3\sqrt2+\sqrt8\)

Show solution

1These aren't like surds yet — simplify \(\sqrt8\) first: \(\sqrt8=\sqrt{4\times2}=2\sqrt2\).
2Now both terms have \(\sqrt2\): \(3\sqrt2+2\sqrt2\)
3\(\boxed{5\sqrt2}\)

Worked example — multiplying Level 2

Simplify: \(\sqrt3\times\sqrt{12}\)

Show solution

1Multiply under one root: \(\sqrt3\times\sqrt{12}=\sqrt{36}\)
2\(\boxed{6}\)

Worked example — dividing Level 2

Simplify: \(\dfrac{\sqrt{48}}{\sqrt3}\)

Show solution

1Combine under one root: \(\dfrac{\sqrt{48}}{\sqrt3}=\sqrt{\dfrac{48}{3}}=\sqrt{16}\)
2\(\boxed{4}\)

Common trap

It's tempting to add surds "straight across," e.g. \(\sqrt2+\sqrt3\stackrel{?}{=}\sqrt5\). Check with decimals: \(\sqrt2+\sqrt3\approx1{,}41+1{,}73=3{,}15\), but \(\sqrt5\approx2{,}24\) — these are not equal. Unlike surds (different numbers under the root) simply cannot be combined into a single surd at all.

Solving Surd Equations

Isolate, square, solve — then ALWAYS check.

Method

Isolate the surd on one side of the equation.
Square both sides to remove the root.
Solve the resulting equation.
Substitute every solution back into the original equation — squaring can introduce a false (extraneous) solution.

Why checking matters

Squaring both sides can turn a false statement into a true one (e.g. \(3=-3\) is false, but squaring both sides gives the true statement \(9=9\)). This means a "solution" produced by squaring might not actually satisfy the original equation — the check is not optional.

Worked example — simplest case Level 1

Solve for \(x\): \(\sqrt{x}=4\)

Show solution

1The surd is already isolated. Square both sides: \(x=16\)
2Check in the original equation: \(\sqrt{16}=4\)✓ valid.
3\(\boxed{x=16}\)

Worked example Level 2

Solve for \(x\): \(\sqrt{x+3}=5\)

Show solution

1The surd is already isolated. Square both sides: \(x+3=25\)
2\(x=22\)
3Check in the original equation: \(\sqrt{22+3}=\sqrt{25}=5\)✓ valid.
4\(\boxed{x=22}\)

Worked example — isolate the surd first Level 3

Solve for \(x\): \(\sqrt{2x-1}-3=0\)

Show solution

1The surd isn't alone yet — add \(3\) to both sides first: \(\sqrt{2x-1}=3\)
2Now square both sides: \(2x-1=9\)
3\(2x=10\Rightarrow x=5\)
4Check in the original equation: \(\sqrt{2(5)-1}-3=\sqrt9-3=3-3=0\)✓ valid
5\(\boxed{x=5}\)

Worked example — catching an extraneous root Level 4

Solve for \(x\): \(\sqrt{x+7}=x-5\)

Show solution

1Square both sides: \(x+7=(x-5)^2=x^2-10x+25\)
2Rearrange to zero: \(x^2-11x+18=0\)
3Factorise: \((x-9)(x-2)=0\Rightarrow x=9\) or \(x=2\)
4Check \(x=9\): \(\sqrt{16}=4\) and \(9-5=4\)✓ valid.
5Check \(x=2\): \(\sqrt9=3\) but \(2-5=-3\). \(3\neq-3\)✗ — reject this extraneous solution.
6\(\boxed{x=9}\) only

Quick Check

Simplify: \(4\sqrt3-\sqrt{27}\)

Why must you check solutions to a surd equation in the ORIGINAL equation?

Completing the Square

Rewrite a quadratic as a perfect square plus a constant.

Quick revision: not every quadratic factorises easily

You already know some quadratics factorise directly, e.g. \(x^2+5x+6=(x+2)(x+3)=0\Rightarrow x=-2\text{ or }x=-3\). Completing the square is a method that works even when no such nice integer factor pair exists.

Method for \(x^2+bx+c=0\)

Move the constant to the other side: \(x^2+bx=-c\)

Add \(\left(\dfrac{b}{2}\right)^2\) to both sides.

Write the left side as a perfect square: \(\left(x+\dfrac{b}{2}\right)^2=\ldots\)

Take the square root of both sides (don't forget \(\pm\)), then solve for \(x\).

Worked example Level 1

Solve for \(x\) by completing the square: \(x^2+4x-5=0\)

Show solution

1Move the constant: \(x^2+4x=5\)
2Half the coefficient of \(x\), then square it: \(\left(\dfrac42\right)^2=4\). Add \(4\) to both sides: \(x^2+4x+4=5+4=9\)
3Write the left side as a perfect square: \((x+2)^2=9\)
4Take the square root of both sides: \(x+2=\pm3\)
5\(\boxed{x=1\text{ or }x=-5}\)

Worked example Level 2

Solve for \(x\) by completing the square: \(x^2+6x+5=0\)

Show solution

1Move the constant: \(x^2+6x=-5\)
2Half the coefficient of \(x\), then square it: \(\left(\dfrac62\right)^2=9\). Add \(9\) to both sides: \(x^2+6x+9=-5+9=4\)
3Write the left side as a perfect square: \((x+3)^2=4\)
4Take the square root of both sides: \(x+3=\pm2\)
5\(\boxed{x=-1\text{ or }x=-5}\)

Don't drop the ±

Forgetting the \(\pm\) in step 4 above and writing only \(x+3=2\) would give just \(x=-1\), silently losing the equally valid solution \(x=-5\).

Worked example — leading coefficient not 1 Level 3

Solve for \(x\) by completing the square: \(3x^2-12x+9=0\)

Show solution

1The method needs a leading coefficient of \(1\), so divide every term by \(3\) first: \(x^2-4x+3=0\)
2Move the constant: \(x^2-4x=-3\)
3Half the coefficient of \(x\), then square it: \(\left(\dfrac{-4}2\right)^2=4\). Add \(4\) to both sides: \(x^2-4x+4=-3+4=1\)
4\((x-2)^2=1\Rightarrow x-2=\pm1\)
5\(\boxed{x=3\text{ or }x=1}\)

The Quadratic Formula

Works every time, even when factorising doesn't.

For \(ax^2+bx+c=0\)

x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}

When to use it

Try factorising first — it's usually faster. Reach for the quadratic formula when the equation doesn't factorise easily with integers, or when you need exact (possibly irrational) roots.

Worked example — leading coefficient 1 Level 1

Solve for \(x\) using the quadratic formula: \(x^2+2x-3=0\)

Show solution

1Identify \(a=1,\,b=2,\,c=-3\).
2Compute the discriminant: \(b^2-4ac=(2)^2-4(1)(-3)=4+12=16\)
3Substitute: \(x=\dfrac{-2\pm\sqrt{16}}{2(1)}=\dfrac{-2\pm4}{2}\)
4\(\boxed{x=1\text{ or }x=-3}\)

Worked example Level 2

Solve for \(x\) using the quadratic formula: \(2x^2-3x-5=0\)

Show solution

1Identify \(a=2,\,b=-3,\,c=-5\).
2Compute \(b^2-4ac\) (this quantity is called the discriminant — you'll meet it properly on the Nature of Roots slide soon): \((-3)^2-4(2)(-5)=9+40=49\)
3Substitute: \(x=\dfrac{-(-3)\pm\sqrt{49}}{2(2)}=\dfrac{3\pm7}{4}\)
4\(\boxed{x=2{,}5\text{ or }x=-1}\)

Note

This same equation factorises too: \(2x^2-3x-5=(2x-5)(x+1)\), giving the identical roots. The formula is a guaranteed fallback, not a replacement for checking if it factorises first.

Worked example — irrational roots Level 3

Solve for \(x\) using the quadratic formula, in simplest surd form: \(x^2-2x-2=0\)

Show solution

1\(a=1,\,b=-2,\,c=-2\). \(b^2-4ac=(-2)^2-4(1)(-2)=4+8=12\)
2\(12\) is not a perfect square, so the roots will be irrational: \(x=\dfrac{2\pm\sqrt{12}}{2}\)
3Simplify the surd, as on the earlier Surd Arithmetic slide: \(\sqrt{12}=\sqrt{4\times3}=2\sqrt3\), so \(x=\dfrac{2\pm2\sqrt3}{2}\)
4\(\boxed{x=1+\sqrt3\text{ or }x=1-\sqrt3}\) (\(\approx2{,}73\) or \(\approx-0{,}73\))

Quick Check

To complete the square on \(x^2+10x+c=0\), what number should replace \(c\) to make the left side a perfect square?

For \(3x^2+x-2=0\), what is the value of the discriminant \(b^2-4ac\)?

Solve for x by completing the square: \(x^2+2x-8=0\)

Quadratic Inequalities

Solve the equation first, then read the sign from the parabola's shape.

Method

Move everything to one side so the inequality compares to zero.
Solve the associated equation to find the critical values (roots).
Since the expression is a parabola, it is positive outside its roots and negative between them when it opens upward (the opposite when it opens downward).

The solid highlighted curve (where \(y=x^2-x-6>0\)) sits outside the roots \(-2\) and \(3\); the dashed muted curve between them is where \(y<0\). Open circles at the roots match the strict inequality used in the worked example.

Worked example — difference of squares Level 1

Solve for \(x\): \(x^2-4\ge0\)

Show solution

1Solve the associated equation: \(x^2-4=0\Rightarrow(x-2)(x+2)=0\Rightarrow x=2\) or \(x=-2\)
2The leading coefficient is positive, so this parabola opens upward — it is positive (or zero) outside its two roots.
3\(\boxed{x\le-2\text{ or }x\ge2}\) (closed, since \(\ge\) includes the roots themselves)

Worked example Level 3

Solve for \(x\): \(x^2-x-6>0\)

Show solution

1Solve the associated equation: \(x^2-x-6=0\Rightarrow(x-3)(x+2)=0\Rightarrow x=3\) or \(x=-2\)
2The leading coefficient is positive, so this parabola opens upward — it is positive (\(>0\)) outside its two roots.
3\(\boxed{x<-2\text{ or }x>3}\)

Open circles at \(-2\) and \(3\) show the endpoints are excluded (\(>\), not \(\ge\)) — the solution is two separate rays, not one segment.

Common trap

Writing only \(x>3\) and forgetting \(x<-2\) is the single most common mark lost on this type of question — always state both rays for a "positive" (outside-the-roots) solution.

Worked example — between the roots Level 3

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

Show solution

1Solve the associated equation: \(x^2-2x-3=0\Rightarrow(x-3)(x+1)=0\Rightarrow x=3\) or \(x=-1\)
2The leading coefficient is positive, so this parabola opens upward — it is negative (\(<0\)) between its two roots this time, the opposite region from the first example above
3\(\boxed{-1

Notice the difference

The first example above (\(>0\), outside the roots) and this one (\(<0\), between the roots) are genuine opposites — both come from the exact same upward-parabola shape, just asking for the positive region versus the negative region. Always check which one the question actually asks for.

Quick Check

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

Why is the solution to a quadratic inequality usually two separate intervals rather than one?

Simultaneous Linear and Quadratic Equations

Substitute the linear equation into the quadratic one.

Method

Rearrange the linear equation into the form \(y=\ldots\) if it isn't already.
Substitute that expression for \(y\) into the quadratic equation.
Solve the resulting quadratic equation for \(x\) (factorise or use the formula).
Substitute each \(x\)-value back into the linear equation (simpler) to find its matching \(y\)-value.

Worked example — simplest case Level 1

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

Show solution

1Both are already equal to \(y\), so set them equal to each other: \(x=x^2\)
2Rearrange to zero: \(x^2-x=0\)
3Factorise: \(x(x-1)=0\Rightarrow x=0\) or \(x=1\)
4Substitute each \(x\)-value into \(y=x\): \(x=0\Rightarrow y=0\); \(x=1\Rightarrow y=1\)
5\(\boxed{(0,0)\text{ and }(1,1)}\)

Worked example Level 3

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

Show solution

1Both are already equal to \(y\), so set them equal to each other: \(x+1=x^2-1\)
2Rearrange to zero: \(x^2-x-2=0\)
3Factorise: \((x-2)(x+1)=0\Rightarrow x=2\) or \(x=-1\)
4Substitute each \(x\)-value into the linear equation (simpler) to find \(y\): \(x=2\Rightarrow y=3\); \(x=-1\Rightarrow y=0\)
5\(\boxed{(2,3)\text{ and }(-1,0)}\)

The two intersection points are exactly the two solutions found algebraically — a line and a parabola can cross at most twice.

Worked example — rearrange first Level 3

Solve simultaneously: \(x-y=1\) and \(y=x^2-3x+2\)

Show solution

1The linear equation isn't in \(y=\ldots\) form yet — rearrange it first: \(x-y=1\Rightarrow y=x-1\)
2Substitute into the quadratic equation: \(x-1=x^2-3x+2\)
3Rearrange to zero: \(0=x^2-4x+3\)
4Factorise: \((x-1)(x-3)=0\Rightarrow x=1\) or \(x=3\)
5Substitute each \(x\)-value into the linear equation \(y=x-1\): \(x=1\Rightarrow y=0\); \(x=3\Rightarrow y=2\)
6\(\boxed{(1,0)\text{ and }(3,2)}\)

Quick Check

Solving \(y=x-1\) and \(y=x^2-3\) simultaneously leads to which equation in \(x\)?

A line and a parabola can intersect at most how many times?

Nature of Roots

The discriminant tells you what kind of roots a quadratic has, without solving it.

The discriminant

\Delta=b^2-4ac

Value of \(\Delta\)	Nature of the roots
\(\Delta>0\)	Two distinct real roots (rational if \(\Delta\) is a perfect square, irrational otherwise)
\(\Delta=0\)	One repeated (equal) real root
\(\Delta<0\)	No real roots

Δ=1>0

Crosses the x-axis twice

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

Δ=0

Touches the x-axis once

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

Δ=-16<0

Never crosses the x-axis

\(x^2+2x+5=0\)

Why this works

The discriminant is exactly the part under the square root in the quadratic formula. If it's negative, there's no real square root to take — that's algebraically why the curve never reaches the x-axis.

Worked example — two distinct rational roots Level 1

Determine the nature of the roots of \(x^2-5x+6=0\), without solving the equation.

Show solution

1\(a=1,\,b=-5,\,c=6\). Compute \(\Delta=(-5)^2-4(1)(6)=25-24=1\)
2\(\Delta>0\), so there are two distinct real roots. \(\Delta=1\) is a perfect square, so \(\boxed{\text{the roots are also rational}}\) (they factorise: \((x-2)(x-3)=0\Rightarrow x=2\text{ or }x=3\))

Worked example — two distinct irrational roots Level 2

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

Show solution

1\(a=1,\,b=-2,\,c=-2\). Compute \(\Delta=(-2)^2-4(1)(-2)=4+8=12\)
2\(\Delta>0\), so there are two distinct real roots. \(12\) is not a perfect square, so \(\boxed{\text{the roots are irrational}}\) (matching the surd-form roots \(1\pm\sqrt3\) found earlier using the quadratic formula)

Worked example Level 2

Determine the nature of the roots of \(x^2+2x+5=0\), without solving the equation.

Show solution

1\(a=1,\,b=2,\,c=5\). Compute \(\Delta=b^2-4ac=4-20=-16\)
2\(\Delta<0\), so \(\boxed{\text{there are no real roots}}\)

Worked example Level 2

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

Show solution

1\(a=1,\,b=-4,\,c=4\). Compute \(\Delta=(-4)^2-4(1)(4)=16-16=0\)
2\(\Delta=0\), so \(\boxed{\text{the roots are real and equal}}\)

Quick Check

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

For \(x^2+x+1=0\), what is the nature of the roots?

For \(x^2-7x+10=0\), what is the nature of the roots?

Grade 11 Mastery Sprint

Work first. Open one answer only when your own line of working is complete.

01 Evaluate: \(27^{\frac23}\)

Answer: \((\sqrt[3]{27})^2=3^2=9\)

02 Solve: \(\sqrt{3x+1}=5\)

Answer: \(3x+1=25\Rightarrow x=8\); check: \(\sqrt{25}=5\).

03 Solve: \(x^2-5x-14=0\)

Answer: \((x-7)(x+2)=0\Rightarrow x=7\text{ or }-2\)

04 Nature of roots: \(x^2+4x+4=0\)

Answer: \(\Delta=4^2-4(1)(4)=0\): real and equal.

Exam Strategy

Checklist before you submit

Surd equation: did you check every solution in the ORIGINAL equation?
Quadratic equation: tried factorising first? If not, is completing the square or the formula genuinely needed?
Quadratic inequality: is your solution TWO intervals, not accidentally just one?
Nature of roots: computed \(\Delta\) correctly before reading off the answer?

Exam mantra

Isolate. Solve. Check. Whichever method you use, isolate what you're solving for, apply the method carefully, and always verify the answer makes sense — especially after squaring.

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What CAPS Expects You to Know

The Grade 11 Algebra & Equations knowledge statement this page is built from.

1
Simplify expressions and solve equations using the laws of exponents for rational exponents, where \(x^{\frac{p}{q}}=\sqrt[q]{x^p}\), \(x>0,\,q>0\).
2
Add, subtract, multiply and divide simple surds; solve simple equations involving surds, checking every solution against the original equation.
3
Solve quadratic equations by completing the square, and by using the quadratic formula.
4
Solve quadratic inequalities in one unknown and interpret the solution graphically.
5
Solve equations in two unknowns, one of which is linear and the other quadratic.
6
Determine the nature of the roots of a quadratic equation, including the case where the roots are equal, using the discriminant.

How to Use This Lesson

A few practical notes before you start.

Always check a surd equation's solutions against the ORIGINAL equation, not the squared version.
Try factorising a quadratic before reaching for completing the square or the formula.
For a quadratic inequality, sketch (even mentally) whether the parabola opens up or down before deciding which side of the roots is the answer.
The discriminant only tells you the NATURE of the roots, not their actual values.
Try each worked example yourself before pressing “Show solution.”

Learn It in Short Videos

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

Rational Exponents

Simplifying expressions with fractional exponents.

Khan Academy · Simplifying quotient of powers (rational exponents)

Surd Arithmetic

Simplifying and combining square roots.

Khan Academy · Simplifying square roots

Completing the Square

Solving a quadratic equation step by step.

Khan Academy · Example 3: Completing the square

The Quadratic Formula

Using the formula to solve any quadratic equation.

Khan Academy · Example 1: Using the quadratic formula

Algebra Study Path

Learn → practise → visualise → stretch.

Step 1 · Learn

Siyavula: Exponents & Surds

Learn exponents, surds and surd equations.

Open the chapter

Step 2 · Practise

Khan Academy: Quadratics

Practise every quadratic method.

Study quadratics

Step 3 · Visualise

GeoGebra Graphing Calculator

See roots, symmetry and inequality regions.

Open GeoGebra

Step 4 · Extend

DBE CAPS Learner Books

Use the official Grade 11 learner book for extra CAPS examples.

Open DBE books

Common Exam Mistakes

Avoid these errors. They cost marks every year.

Not checking surd equation solutions

Squaring can introduce a false solution — always substitute back into the original (unsquared) equation.

Adding surds that aren't like terms

\(\sqrt2+\sqrt3\) cannot be combined into a single surd — only identical surds can be added or subtracted.

Forgetting the ± sign

Both completing the square and the quadratic formula involve a square root — forgetting \(\pm\) loses one of the two solutions.

Giving one interval instead of two

A quadratic inequality's solution is usually two separate intervals (outside or between the roots) — check both sides, not just one.

Solving instead of just stating the nature of roots

If a question only asks for the NATURE of the roots, computing the discriminant is enough — there is no need to solve the equation fully.

Substituting into the harder equation

When finding \(y\) after solving simultaneous linear-quadratic equations, always substitute back into the LINEAR equation — it's simpler and less error-prone.

Practise This Topic

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

BANK

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Grade 11 Algebra & Equations Mastery Bank

Exam-style Grade 11 questions arranged by level, combining original practice with clearly identified paper-and-memo matches.

Step 1 • Practise • Cognitive Levels

Open Mastery Bank

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Algebra & Equations Grade 11 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 11 CAPS questions about algebra and equations.

What is Grade 11 Algebra & Equations about?

Grade 11 Algebra covers rational exponents (simplifying and solving equations with fractional exponents), surd arithmetic (adding, subtracting, multiplying and dividing simple surds), solving equations involving surds, completing the square, the quadratic formula, quadratic inequalities, simultaneous equations where one equation is linear and the other quadratic, and determining the nature of the roots of a quadratic equation using the discriminant. This is the single highest-weighted topic in the entire Grade 11 Paper 1 curriculum.

Why must I check my answer after solving a surd equation?

Squaring both sides of an equation can introduce an extraneous (false) solution that does not actually satisfy the original equation. Always substitute every solution back into the original surd equation to check it is genuinely valid.

When should I use the quadratic formula instead of factorising?

Use the quadratic formula whenever a quadratic equation does not factorise easily with integers — it always works, even when the roots are irrational or when factorising would take too long to spot.

What does the discriminant tell you?

The discriminant, \(b^2-4ac\), tells you the nature of a quadratic equation's roots without solving it: positive means two distinct real roots, zero means one repeated (equal) real root, and negative means no real roots.

Which resource should I open first for this topic?

Everything you need to learn the topic is on this page already. Once you've been through the notes above, work through the Mastery Bank, then finish with the Test Your Knowledge quiz as a self-check.

What mistakes should I avoid?

Common mistakes include forgetting to check for extraneous roots after squaring a surd equation, adding surds that are not like terms, forgetting the plus-or-minus when completing the square or applying the quadratic formula, and forgetting that a quadratic inequality's solution is usually two separate intervals, not one.

Algebra & Equations — Grade 11

Algebra & Equations

The 3-Year Algebra Journey

The Biggest Topic This Year

Rational Exponents

Adding, Subtracting and Multiplying Surds

Solving Surd Equations

Completing the Square

The Quadratic Formula

Quadratic Inequalities

Simultaneous Linear and Quadratic Equations

Nature of Roots

Grade 11 Mastery Sprint

Exam Strategy

What CAPS Expects You to Know

How to Use This Lesson

Learn It in Short Videos

Rational Exponents

Surd Arithmetic

Completing the Square

The Quadratic Formula

Algebra Study Path

Siyavula: Exponents & Surds

Khan Academy: Quadratics

GeoGebra Graphing Calculator

DBE CAPS Learner Books

Common Exam Mistakes

Not checking surd equation solutions

Adding surds that aren't like terms

Forgetting the ± sign

Giving one interval instead of two

Solving instead of just stating the nature of roots

Substituting into the harder equation

This Topic in Other Grades

Grade 10

Grade 12

Practise This Topic

Frequently Asked Questions

What is Grade 11 Algebra & Equations about?

Why must I check my answer after solving a surd equation?

When should I use the quadratic formula instead of factorising?

What does the discriminant tell you?

Which resource should I open first for this topic?

What mistakes should I avoid?