2.4 Solving quadratic equations, including in a function of the unknown — Practise

Work through these on your own. Give surd answers exactly. When an equation is quadratic in a function of the unknown, write down the substitution you choose before you solve.

Both the questions and the answers below come from a set that is checked by computer every time the book is built.

Solve $x^2 - 5x + 6 = 0$ .
Solve $2x^2 + 5x - 3 = 0$ .
Solve $x^2 + 4x - 1 = 0$ , giving the answer in surd form.
Solve $x^2 - 3x - 2 = 0$ , giving the answer in surd form.
Solve $x^4 - 5x^2 + 4 = 0$ .
Solve $2^{2x} - 5 \times 2^x + 4 = 0$ .
Solve $x + 3\sqrt{x} - 10 = 0$ .
Solve $\dfrac{1}{x^2} - \dfrac{3}{x} + 2 = 0$ , where $x \neq 0$ .
Solve $x^6 - 9x^3 + 8 = 0$ for real $x$ .
Solve $3x^2 = 12$ .
Solve $(x - 1)^2 = 9$ .
Solve $\sqrt{x + 5} = x - 1$ , and explain why one of the values found by squaring is rejected.

Answers and concise solutions

1. The solutions

Answer: Factorise: $x^2 - 5x + 6 = (x - 2)(x - 3)$ . A product is zero when a factor is zero, so $x = 2$ or $x = 3$ .

2. The solutions

Answer: Split the middle term: $2x^2 + 6x - x - 3 = (2x - 1)(x + 3)$ . So $x = \frac{1}{2}$ or $x = -3$ . Alternatively, the formula with $a = 2$ , $b = 5$ and $c = -3$ gives the same two roots.

3. The solutions

Answer: Complete the square: $x^2 + 4x - 1 = (x + 2)^2 - 5$ , so $(x + 2)^2 = 5$ . Then $x + 2 = \pm\sqrt{5}$ , so $x = -2 \pm \sqrt{5}$ .

4. The solutions

Answer: Use the formula with $a = 1$ , $b = -3$ and $c = -2$ : $x = \dfrac{3 \pm \sqrt{9 + 8}}{2} = \dfrac{3 \pm \sqrt{17}}{2}$ . Neither root factorises, so the surd form is the exact answer.

5. The solutions

Answer: Let $t = x^2$ . Then $t^2 - 5t + 4 = 0$ , so $(t - 1)(t - 4) = 0$ and $t = 1$ or $t = 4$ . Replace $t$ : $x^2 = 1$ gives $x = \pm 1$ , and $x^2 = 4$ gives $x = \pm 2$ . There are four solutions.

6. The solutions

Answer: Let $t = 2^x$ . Then $t^2 - 5t + 4 = 0$ , so $t = 1$ or $t = 4$ . Replace $t$ : $2^x = 1$ gives $x = 0$ , and $2^x = 4$ gives $x = 2$ .

7. The solution

Answer: Let $t = \sqrt{x}$ , so $x = t^2$ and $t \geq 0$ . Then $t^2 + 3t - 10 = 0$ , so $(t + 5)(t - 2) = 0$ and $t = 2$ or $t = -5$ . The value $t = -5$ is rejected because a square root is never negative. So $\sqrt{x} = 2$ and $x = 4$ . Check: $4 + 3 \times 2 - 10 = 0$ .

8. The solutions

Answer: Let $t = \frac{1}{x}$ . Then $t^2 - 3t + 2 = 0$ , so $(t - 1)(t - 2) = 0$ and $t = 1$ or $t = 2$ . Replace $t$ : $\frac{1}{x} = 1$ gives $x = 1$ , and $\frac{1}{x} = 2$ gives $x = \frac{1}{2}$ . Neither value is zero, so both are allowed.

9. The real solutions

Answer: Let $t = x^3$ . Then $t^2 - 9t + 8 = 0$ , so $(t - 1)(t - 8) = 0$ and $t = 1$ or $t = 8$ . Replace $t$ : $x^3 = 1$ gives $x = 1$ , and $x^3 = 8$ gives $x = 2$ . A cube has exactly one real root, so these are the only real solutions.

10. The solutions

Answer: Divide both sides by $3$ : $x^2 = 4$ . Taking both signs of the square root gives $x = 2$ or $x = -2$ . Remember the negative root.

11. The solutions

Answer: A square equals $9$ when the bracket is $3$ or $-3$ . So $x - 1 = 3$ gives $x = 4$ , and $x - 1 = -3$ gives $x = -2$ . Do not take only the positive square root.

12. The solution

Answer: Square both sides: $x + 5 = x^2 - 2x + 1$ , so $x^2 - 3x - 4 = 0$ and $(x - 4)(x + 1) = 0$ . The values are $x = 4$ and $x = -1$ . Check the original equation: for $x = 4$ , $\sqrt{9} = 3$ and $4 - 1 = 3$ , so $x = 4$ works. For $x = -1$ , $\sqrt{4} = 2$ but $-1 - 1 = -2$ , so $x = -1$ fails. It is an extraneous root created by squaring, and the only solution is $x = 4$ .