Solve the quadratic equation $x^{2} - 3 x - 9 = 0$ for x and give your answer correct to two decimal places.

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Solution

Quadratic formula
Compare $x^{2} - 3 x - 9 = 0$ with $a x^{2} + b x + c = 0,$ we get $a = 1, b = - 3, c = - 9$
$\begin{array}{rcl} x & = & \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a} \\ \Rightarrow & x = \frac{- (- 3) \pm \sqrt{{(- 3)}^{2} - 4 (1) (- 9)}}{2 (1)} \\ \Rightarrow & x = \frac{3 \pm \sqrt{9 + 36}}{2} \\ \Rightarrow & x = \frac{3 \pm \sqrt{45}}{2} \\ \Rightarrow & x = \frac{3 \pm 6.70}{2} \\ \Rightarrow & x = \frac{9.70}{2}, \frac{- 3.70}{2} \\ x & = & 4.85, - 1.85 \end{array}$
Hence, the required solution for x are $4.85, - 1.85 .$

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