Solving Quadratic Equations

Q: How do I know when to use k-substitution?

When the same expression, such as $x^2 - 2x$ or $2^x$, appears more than once in the equation.

Quadratic equations open almost every Paper 1. The algebra section is worth about 25 marks, and the same skills are used again in functions, calculus and finance, so speed and accuracy here pay off across the whole paper.

Solving by factorising

Write the equation in standard form $a x^{2} + b x + c = 0$ , factorise, then use the zero product rule: if $A B = 0$ , then $A = 0$ or $B = 0$ .

Example. $x^{2} - x - 12 = 0$

$(x - 4) (x + 3) = 0 \Rightarrow x = 4 or x = - 3$

Never divide both sides by $x$ or by a bracket that contains $x$ . You lose a solution. Factorise it out instead: $x^{2} = 5 x$ gives $x (x - 5) = 0$ , so $x = 0$ or $x = 5$ .

Solving with the quadratic formula

If the quadratic does not factorise easily, use

$x = \frac{- b \pm b ^{2} - 4 a c}{2 a}$

Example. $2 x^{2} - 3 x - 4 = 0$ , so $a = 2$ , $b = - 3$ , $c = - 4$ .

$x = \frac{3 \pm 9 + 32}{4} = \frac{3 \pm 41}{4}$

$x \approx 2, 35$ or $x \approx - 0, 85$

Put brackets around negative values of $b$ when you substitute. Give answers in the form asked: simplest surd form or rounded to two decimal places.

Solving by substitution (k-method)

If the same expression appears more than once, replace it with $k$ to get a simpler quadratic.

Example. $(x^{2} - 2 x)^{2} - 11 (x^{2} - 2 x) + 24 = 0$

Let $k = x^{2} - 2 x$ :

$k^{2} - 11 k + 24 = 0 \Rightarrow (k - 3) (k - 8) = 0$

$x^{2} - 2 x = 3 \Rightarrow (x - 3) (x + 1) = 0 \Rightarrow x = 3$ or $x = - 1$
$x^{2} - 2 x = 8 \Rightarrow (x - 4) (x + 2) = 0 \Rightarrow x = 4$ or $x = - 2$

Equations with fractions. Multiply by the lowest common denominator, after stating restrictions (values that make a denominator 0). Check that no answer breaks a restriction.

Any correct method may be used to solve a quadratic: factorising, completing the square or the quadratic formula. Completing the square is also needed for circles and turning points.

Key terms

Standard form: $a x^{2} + b x + c = 0$ , with everything on one side.
Zero product rule: If $A B = 0$ , then $A = 0$ or $B = 0$ .
Restriction: A value the variable cannot take, usually because a denominator would be 0.
k-substitution: Replacing a repeated expression with a single variable to simplify an equation.

Common mistakes

Dividing by $x$ or by a factor with $x$ in it, losing a solution.
Not writing the equation in standard form before factorising.
Substituting $- b$ incorrectly when $b$ is negative.
Stopping after finding $k$ and not solving for $x$ .
Not checking answers against restrictions in fraction equations.

Practice questions

Reveal steps one at a time, then tick the marks you earned.

Question 1

3 marks total

knowledge

Solve for $x$ : $x^{2} - x - 12 = 0$

Worked solution

Hidden - try the question first

Solution

$(x - 4) (x + 3) = 0$
$x = 4$
or $x = - 3$

Question 2

5 marks total

routine

Solve for $x$ , correct to two decimal places: $2 x^{2} - 3 x - 4 = 0$

Worked solution

Hidden - try the question first

Solution

$(x - 4) (x + 3) = 0$
$= \frac{3 \pm 41}{4}$
$x \approx 2, 35$
or $x \approx - 0, 85$

Question 3

6 marks total

complex

Solve for $x$ : $(x^{2} - 2 x)^{2} - 11 (x^{2} - 2 x) + 24 = 0$

Worked solution

Hidden - try the question first

Solution

Let $k = x^{2} - 2 x$ : $k^{2} - 11 k + 24 = 0$
$(k - 3) (k - 8) = 0$
$x^{2} - 2 x - 3 = 0 \Rightarrow (x - 3) (x + 1) = 0$
$x = 3$ or $x = - 1$
$x^{2} - 2 x - 8 = 0 \Rightarrow (x - 4) (x + 2) = 0$
$x = 4$ or $x = - 2$

Question 4

8 marks total

problem-solving

Solve for $x$ : $\frac{3}{x - 1} - \frac{2}{x + 2} = 1$

Worked solution

Hidden - try the question first

Solution

Restrictions: $k = x^{2} - 2 x$ and $x \neq = - 2$
Multiply by $(x - 1) (x + 2)$ : $3 (x + 2) - 2 (x - 1) = (x - 1) (x + 2)$
$x + 8 = x^{2} + x - 2$
$x^{2} = 10$
$x = \pm 10$
Neither value breaks a restriction

FAQs

Can I solve quadratics by completing the square in IEB Grade 12?

Yes, any correct method earns the marks. Most learners use factorising or the formula. You also need completing the square to find turning points and to rewrite circle equations.

When should I use the quadratic formula instead of factorising?

If you cannot find factors within about 30 seconds, or the question asks for decimal or surd answers, use the formula.

How do I know when to use k-substitution?

When the same expression, such as $x^{2} - 2 x$ or $2^{x}$ , appears more than once in the equation.

Solving by factorising

Solving with the quadratic formula

Solving by substitution (k-method)

Key terms

Common mistakes

Practice questions

FAQs

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