Quadratic Equations | Exercise 4.2

Question 1

Find the roots of the following quadratic equations by factorisation:

(i) $x^2 - 3x - 10 = 0$

(ii) $2x^2 + x - 6 = 0$

(iii) $\sqrt{2}x^2 + 7x + 5\sqrt{2} = 0$

(iv) $2x^2 - x + \dfrac{1}{8} = 0$

(v) $100x^2 - 20x + 1 = 0$

Solution

Understand the Question

To find the roots of a quadratic equation $ax^2 + bx + c = 0$ by factorisation (splitting the middle term):

Find two numbers that multiply to $a \times c$ and add up to $b$ .
Split the middle term $bx$ using these two numbers and factor by grouping.
Set each linear factor equal to zero using the zero-product property to find the values of $x$ (the roots).

(i) $x^2 - 3x - 10 = 0$

Step 1 · Factorise and Find Roots

Find two numbers whose product is $1 \times (-10) = -10$ and sum is $-3$ . The numbers are $-5$ and $2$ .

\begin{aligned} x^2 - 5x + 2x - 10 &= 0 \\ x(x - 5) + 2(x - 5) &= 0 \\ (x - 5)(x + 2) &= 0 \end{aligned}

Setting each factor to zero

\begin{aligned} x - 5 = 0 \quad &\text{or} \quad x + 2 = 0 \\ x = 5 \quad &\text{or} \quad x = -2 \end{aligned}

Answer

(i) $x = 5, -2$

(ii) $2x^2 + x - 6 = 0$

Step 1 · Factorise and Find Roots

Find two numbers whose product is $2 \times (-6) = -12$ and sum is $-1$ . The numbers are $-4$ and $3$ .

\begin{aligned} 2x^2 - 4x + 3x - 6 &= 0 \\ 2x(x - 2) + 3(x - 2) &= 0 \\ (x - 2)(2x + 3) &= 0 \end{aligned}

Setting each factor to zero

\begin{aligned} x - 2 = 0 \quad &\text{or} \quad 2x + 3 = 0 \\[0.6em] x = 2 \quad &\text{or} \quad 2x = -3 \\[0.6em] x = 2 \quad &\text{or} \quad x = -\dfrac{3}{2} \end{aligned}

Answer

(ii) $x = 2, -\dfrac{3}{2}$

(iii) $\sqrt{2}x^2 + 7x + 5\sqrt{2} = 0$

Step 1 · Factorise and Find Roots

Find two numbers whose product is $\sqrt{2} \times 5\sqrt{2} = 10$ and sum is $7$ . The numbers are $5$ and $2$ .

\begin{aligned} \sqrt{2}x^2 + 5x + 2x + 5\sqrt{2} &= 0 \\ x(\sqrt{2}x + 5) + \sqrt{2}(\sqrt{2}x + 5) &= 0 \\ (\sqrt{2}x + 5)(x + \sqrt{2}) &= 0 \end{aligned}

Setting each factor to zero

\begin{aligned} \sqrt{2}x + 5 = 0 \quad &\text{or} \quad x + \sqrt{2} = 0 \\[0.6em] \sqrt{2}x = -5 \quad &\text{or} \quad x = -\sqrt{2} \\[0.6em] x = -\dfrac{5}{\sqrt{2}} \quad &\text{or} \quad x = -\sqrt{2} \end{aligned}

Answer

(iii) $x = -\dfrac{5}{\sqrt{2}}, -\sqrt{2}$

(iv) $2x^2 - x + \dfrac{1}{8} = 0$

Step 1 · Factorise and Find Roots

Multiply the entire equation by $8$ to clear the fraction $16x^2 - 8x + 1 = 0$

Find two numbers whose product is $16 \times 1 = 16$ and sum is $-8$ . The numbers are $-4$ and $-4$ .

\begin{aligned} 16x^2 - 4x - 4x + 1 &= 0 \\ 4x(4x - 1) - 1(4x - 1) &= 0 \\ (4x - 1)(4x - 1) &= 0 \end{aligned}

Setting each factor to zero

\begin{aligned} 4x - 1 = 0 \quad &\text{or} \quad 4x - 1 = 0 \\[0.6em] 4x = 1 \quad &\text{or} \quad 4x = 1 \\[0.6em] x = \dfrac{1}{4} \quad &\text{or} \quad x = \dfrac{1}{4} \end{aligned}

Answer

(iv) $x = \dfrac{1}{4}, \dfrac{1}{4}$

(v) $100x^2 - 20x + 1 = 0$

Step 1 · Factorise and Find Roots

Find two numbers whose product is $100 \times 1 = 100$ and sum is $-20$ . The numbers are $-10$ and $-10$ .

\begin{aligned} 100x^2 - 10x - 10x + 1 &= 0 \\ 10x(10x - 1) - 1(10x - 1) &= 0 \\ (10x - 1)(10x - 1) &= 0 \end{aligned}

Setting each factor to zero

\begin{aligned} 10x - 1 = 0 \quad &\text{or} \quad 10x - 1 = 0 \\[0.6em] 10x = 1 \quad &\text{or} \quad 10x = 1 \\[0.6em] x = \dfrac{1}{10} \quad &\text{or} \quad x = \dfrac{1}{10} \end{aligned}

Answer

(v) $x = \dfrac{1}{10}, \dfrac{1}{10}$

Common Mistakes

Sign Errors in Middle-Term Splitting: Selecting factors that give the right magnitude but incorrect signs (e.g., choosing $+5$ and $-2$ instead of $-5$ and $+2$ for a sum of $-3$ ).
Radical Factoring Confusion: In equations with square roots like $\sqrt{2}x^2 + 7x + 5\sqrt{2} = 0$ , forgetting that $2x = \sqrt{2} \times \sqrt{2}x$ , which is essential for factoring by grouping.
Omitting Repeated Roots: When both linear factors are identical (e.g. $(4x - 1)^2 = 0$ ), write both identical roots ( $x = \dfrac{1}{4}, \dfrac{1}{4}$ ) since a quadratic equation always has two roots.

More questions in Exercise 4.2