Equations in Quadratic Form | IEB Grade 11 Maths

Some equations are not quadratic at first sight but become quadratic after a substitution or after clearing fractions. Recognising these patterns turns a hard-looking question into routine work.

Substitution to reduce to a quadratic

If the same expression appears squared and on its own, replace it with $k$ .

Example. $x^{4} - 5 x^{2} + 4 = 0$ . Let $k = x^{2}$ :

$k^{2} - 5 k + 4 = 0 \Rightarrow (k - 1) (k - 4) = 0$

$x^{2} = 1$ gives $x = \pm 1$ , and $x^{2} = 4$ gives $x = \pm 2$ .

Example. $(x^{2} + x)^{2} - 8 (x^{2} + x) + 12 = 0$ . Let $k = x^{2} + x$ : $k = 2$ or $k = 6$ . Then $x^{2} + x - 2 = 0$ gives $x = 1$ or $x = - 2$ , and $x^{2} + x - 6 = 0$ gives $x = 2$ or $x = - 3$ .

Always substitute back and solve for $x$ .

Equations with fractions leading to quadratics

State the restrictions: values that make any denominator 0.
Multiply every term by the lowest common denominator.
Solve the quadratic.
Reject any answer that breaks a restriction.

Example. $\frac{2}{x - 1} + \frac{3}{x + 1} = 2$ , with $x \neq = \pm 1$ .

$2 (x + 1) + 3 (x - 1) = 2 (x^{2} - 1) \Rightarrow 2 x^{2} - 5 x - 1 = 0$

$x = \frac{5 \pm 33}{4} \Rightarrow x \approx 2, 69 or x \approx - 0, 19$

Multiply every term, including those without a denominator.

Restrictions on the variable

Restrictions come from:

denominators: cannot be 0
square roots: the expression under the root cannot be negative, and $\dots \geq 0$

Example. $x + 2 x - 8 = 0$ . Let $k = x$ , with $k \geq 0$ :

$k^{2} + 2 k - 8 = 0 \Rightarrow (k + 4) (k - 2) = 0$

$k = - 4$ is impossible, so $x = 2$ and $x = 4$ .

Key terms

Substitution: Replacing a repeated expression with a single letter.
Restriction: A value the variable may not take.
Lowest common denominator: The smallest expression that every denominator divides into.

Common mistakes

Stopping at the values of $k$ .
Missing the negative roots when $x^{2} = k$ .
Not multiplying every term by the LCD.
Keeping an answer that makes a denominator zero.

Practice questions

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

Question 1

4 marks total

routine

Solve for $x$ : $x^{4} - 5 x^{2} + 4 = 0$

Worked solution

Hidden - try the question first

Solution

Let $k = x^{2}$ : $k^{2} - 5 k + 4 = 0$
$(k - 1) (k - 4) = 0$
$x^{2} = 1 \Rightarrow x = \pm 1$
$x^{2} = 4 \Rightarrow x = \pm 2$

Question 2

5 marks total

routine

Solve for $x$ : $(x^{2} + x)^{2} - 8 (x^{2} + x) + 12 = 0$

Worked solution

Hidden - try the question first

Solution

Let $k = x^{2}$ : $k^{2} - 8 k + 12 = 0$
$(k - 2) (k - 6) = 0$
$x^{2} + x - 2 = 0 \Rightarrow x = 1$ or $x = - 2$
$x^{2} + x - 6 = 0 \Rightarrow (x + 3) (x - 2) = 0$
$x = 2$ or $x = - 3$

Question 3

6 marks total

complex

Solve for $x$ , correct to two decimal places: $\frac{2}{x - 1} + \frac{3}{x + 1} = 2$

Worked solution

Hidden - try the question first

Solution

Restrictions: $k = x^{2}$ and $x \neq = - 1$
$2 (x + 1) + 3 (x - 1) = 2 (x - 1) (x + 1)$
$5 x - 1 = 2 x^{2} - 2 \Rightarrow 2 x^{2} - 5 x - 1 = 0$
$x = \frac{5 \pm 33}{4}$
$x \approx 2, 69$ or $x \approx - 0, 19$
Neither breaks a restriction

Question 4

8 marks total

problem-solving

Solve for $x$ :

(a) $x + \frac{6}{x} = 5$ (3)

(b) $x + 2 x - 8 = 0$ (5)

Worked solution

Hidden - try the question first

Solution

(a) $k = x^{2}$ ; $x \neq = - 1$
$(x - 2) (x - 3) = 0$
$x = 2$ or $x = 3$
(b) Let $k = x$ , where $k \geq 0$
$x^{2} + x - 6 = 0 \Rightarrow (x + 3) (x - 2) = 0$
$(k + 4) (k - 2) = 0$
$k = - 4$ is rejected because $x \geq 0$
$x = 2 \Rightarrow x = 4$

FAQs

How do I spot an equation in quadratic form?

Look for an expression that appears once squared and once on its own, such as $x^{2}$ and $x^{4}$ , or $(x^{2} + x)^{2}$ and $(x^{2} + x)$ .

Why do I need restrictions?

Multiplying by an expression that could be 0 can introduce false solutions. Restrictions show which answers to reject.

Is k-substitution in Grade 12 too?

Yes. It is examinable in Grade 12, including with exponential expressions such as $2^{x}$ .

Equations in quadratic form (k-substitution)

Substitution to reduce to a quadratic

Equations with fractions leading to quadratics

Restrictions on the variable

Key terms

Common mistakes

Practice questions

FAQs

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