Question

How do you use the quadratic formula to find the roots of the equation ${{x}^{2}}-6x-19=0$ ?

Answer 1

For answering this question we need to find the roots of the equation ${{x}^{2}}-6x-19=0$ using the quadratic formula. From the concepts we know that the quadratic formula states that for any quadratic equation in the form of $a{{x}^{2}}+bx+c=0$ the roots are given by the formulae $\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ .

Complete step-by-step solution:
Now considering from the question we have been asked to find the roots of the equation ${{x}^{2}}-6x-19=0$ using the quadratic formula.
From the basics of concepts we know that the quadratic formula states that for any quadratic equation in the form of $a{{x}^{2}}+bx+c=0$ the roots are given by the formulae $\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ .
By using this formula here we can find the roots of the given equation.
After applying we will have
$\begin{aligned} & \Rightarrow \dfrac{-\left( -6 \right)\pm \sqrt{{{\left( -6 \right)}^{2}}-4\left( 1 \right)\left( -19 \right)}}{2\left( 1 \right)} \\\ & \Rightarrow \dfrac{6\pm \sqrt{36+76}}{2} \\\ & \Rightarrow \dfrac{6\pm \sqrt{112}}{2} \\\ & \Rightarrow \dfrac{6\pm \sqrt{4\times 4\times 7}}{2} \\\ & \Rightarrow \dfrac{6\pm 4\sqrt{7}}{2} \\\ & \Rightarrow 3\pm 2\sqrt{7} \\\ \end{aligned}$
Therefore we can conclude that the roots of the equation ${{x}^{2}}-6x-19=0$ is given as
$3\pm 2\sqrt{7}$.

Note: In questions of this type we should be sure with the concepts and the formulae. The calculations that we perform during the solution of this question should be done carefully and should be sure with them. The verification of the answers we find can be done by multiplying the factors of the equation that are obtained from the roots of the equation. So now the factors of the equation ${{x}^{2}}-6x-19=0$ are $\left( x-3\pm 2\sqrt{7} \right)$ obtained from the roots of the equation which are $3\pm 2\sqrt{7}$ . By multiplying the factors we will have

$\begin{aligned} & \left( x-3-2\sqrt{7} \right)\left( x-3+2\sqrt{7} \right) \\\ & \Rightarrow {{\left( x-3 \right)}^{2}}-{{\left( 2\sqrt{7} \right)}^{2}} \\\ & \Rightarrow {{x}^{2}}+9-6x-28 \\\ & \Rightarrow {{x}^{2}}-6x-19 \\\ \end{aligned}$ .