Solve the quadratic equation, x² - 3(x + 3) = 0; giving your answer correct to two significant figures.

Given,

⇒ x² - 3(x + 3) = 0

⇒ x² - 3x - 9 = 0

Comparing x² - 3x - 9 = 0 with ax² + bx + c = 0 we get,

a = 1, b = -3 and c = -9.

We know that,

x = $\dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Substituting values of a, b and c in above equation we get,

$\Rightarrow x = \dfrac{-(-3) \pm \sqrt{(-3)^2 - 4.(1).(-9)}}{2(1)} \\[1em] = \dfrac{3 \pm \sqrt{9 + 36}}{2} \\[1em] = \dfrac{3 \pm \sqrt{45}}{2} \\[1em] = \dfrac{3 \pm 6.7}{2} \\[1em] = \dfrac{3 + 6.7}{2} \text{ or } \dfrac{3 - 6.7}{2} \\[1em] = \dfrac{9.7}{2} \text{ or } \dfrac{-3.7}{2} \\[1em] = 4.9 \text{ or } -1.9$

Hence, x = 4.9 or -1.9