Solve the quadratic equation $(x - 2)^2 + 11x - 7$ = 0 and give your answer correct to 3 significant figures.
Show SolutionHide Solution↓
Given,
$$\begin{aligned}\Rightarrow (x - 2)^2 + 11x - 7 = 0 \\[1em] \Rightarrow x^2 - 4x + 4 + 11x - 7 = 0 \\[1em] \Rightarrow x^2 + 7x - 3 = 0\end{aligned}$$
Comparing the above equation with $ax^2 + bx + c$ = 0, we get :
$a = 1, b = 7 \text{ and } c = -3$
By quadratic formula,
$$\begin{aligned}\Rightarrow x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \\[1em] \Rightarrow x = \frac{-7 \pm \sqrt{7^2 - 4 \times 1 \times (-3)}}{2 \times 1} \\[1em] \Rightarrow x = \frac{-7 \pm \sqrt{49 + 12}}{2} \\[1em] \Rightarrow x = \frac{-7 \pm \sqrt{61}}{2} \\[1em] \Rightarrow x = \frac{-7 \pm 7.8102}{2}\end{aligned}$$
Solving further,
$$\begin{aligned}\Rightarrow x = \frac{-7 + 7.8102}{2} \text{ or } x = \frac{-7 - 7.8102}{2} \\[1em] \Rightarrow x = \frac{0.8102}{2} \text{ or } x = \frac{-14.8102}{2} \\[1em] \Rightarrow x = 0.4051 \text{ or } x = -7.4051\end{aligned}$$
Correct to 3 significant figures,
$x = 0.405 \text{ or } x = -7.41$.
Hence, $\bm{x}$ = 0.405 or $\bm{x}$ = −7.41.