Question
Answer Key
Worked Solution
9758/2021/P1/Q02
(a) Sketch, on the same axes, the graphs of $y=\left|x^2-5\right|$ and $y=2 x-1$.
[2]
(b) Find the exact solutions of $\left|x^2-5\right|=2 x-1$.
[4]
![](https://sgmathsacad.com/wp-content/uploads/2022/10/image-52.png)
(b) $x=1+\sqrt{5}$ or $x=-1+\sqrt{7}$
![](https://sgmathsacad.com/wp-content/uploads/2022/10/image-52.png)
$\left|x^2-5\right|=2 x-1$
$x^2-5 =2 x-1$
$x^2-2 x-4 =0$
$x =\frac{2 \pm \sqrt{(-2)^2-4(1)(-4)}}{2(1)}$
$=\frac{2 \pm \sqrt{20}}{2}$
$=\frac{2 \pm 2 \sqrt{5}}{2}$
$=1 \pm \sqrt{5}$
or
$x^2-5 =-(2 x-1)$
$x^2-5+2 x-1 =0$
$x^2+2 x-6 =0$
$x =\frac{-2 \pm \sqrt{(2)^2-4(1)(-6)}}{2(1)}$
$=\frac{-2 \pm \sqrt{28}}{2}$
$=\frac{-2 \pm 2 ; \overline{7}}{2}$
$=-1 \pm i \overline{7}$
Since $\left|x^2-5\right|=2 x-1 \geqslant 0$
$\Rightarrow x \geqslant \frac{1}{2}$
$\therefore x=1+\sqrt{5} \quad$ or $\quad x=-1+\sqrt{7}$