Question

Solve the following quadratic equation and give your answer correct to 3 significant figures:
$5x^2-3x-4=0$.

Answer 1

We know that the roots of a quadratic equation $a{x}^2+bx+c=0$ can be calculated using quadratic formula,
$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

On comparing the given equation with standard form, we get $a=5,b=-3$, and $c=-4$.

On substituting the values in the formula, we get:
$\begin{array}{cc}x& =\frac{3\pm \sqrt{(-3{)}^2-4\times 5\times (-4)}}{2\times 5}\\ & =\frac{3\pm \sqrt{9+80}}{10}\\ & =\frac{3\pm 9.433}{10}\\ \Rightarrow \text{either}x& =\frac{3+9.433}{10}\text{or}x=\frac{3-9.433}{10}\\ \Rightarrow x& =\frac{12.433}{10}\text{or}x=-\frac{6.433}{10}\\ \Rightarrow x& =1.243\text{or}x=-0.643\end{array}$

Hence, the roots of the given equation correct to 3 significant figures are $x=1.243$ or $-0.643$.