Question

How do you solve $-5{{x}^{2}}-8x+1=0$ using the quadratic formula?

Answer 1

Hint: We start solving the problem by recalling the quadratic formula as the roots of the quadratic equation $a{{x}^{2}}+bx+c=0$ is defined as $\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$. We then compare the given equation with the $a{{x}^{2}}+bx+c=0$ to get the values of a, b and c. We then substitute these values in the quadratic formula $\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ and then make the necessary calculations to get the required value of roots of the given equation.

Complete step by step answer:
According to the problem, we are asked to solve the given equation $-5{{x}^{2}}-8x+1=0$ using the quadratic formula.
We have given the equation $-5{{x}^{2}}-8x+1=0$ ---(1).
Let us recall the quadratic formula. We know that the roots of the quadratic equation $a{{x}^{2}}+bx+c=0$ is defined as $\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$. Let us use this result in equation (1).
Comparing $-5{{x}^{2}}-8x+1=0$ with $a{{x}^{2}}+bx+c=0$, we get $a=-5$, $b=-8$ and $c=1$. Now, let us substitute these values in $\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ to get the roots of the equation $-5{{x}^{2}}-8x+1=0$.
Now, we have $\dfrac{-\left( -8 \right)\pm \sqrt{{{\left( -8 \right)}^{2}}-4\left( -5 \right)\left( 1 \right)}}{2\left( -5 \right)}=\dfrac{8\pm \sqrt{64+20}}{-10}$.
$\Rightarrow \dfrac{-\left( -8 \right)\pm \sqrt{{{\left( -8 \right)}^{2}}-4\left( -5 \right)\left( 1 \right)}}{2\left( -5 \right)}=\dfrac{8\pm \sqrt{84}}{-10}$.
$\Rightarrow \dfrac{-\left( -8 \right)\pm \sqrt{{{\left( -8 \right)}^{2}}-4\left( -5 \right)\left( 1 \right)}}{2\left( -5 \right)}=\dfrac{8\pm \sqrt{4\times 21}}{-10}$.
$\Rightarrow \dfrac{-\left( -8 \right)\pm \sqrt{{{\left( -8 \right)}^{2}}-4\left( -5 \right)\left( 1 \right)}}{2\left( -5 \right)}=\dfrac{8\pm 2\sqrt{21}}{-10}$.
$\Rightarrow \dfrac{-\left( -8 \right)\pm \sqrt{{{\left( -8 \right)}^{2}}-4\left( -5 \right)\left( 1 \right)}}{2\left( -5 \right)}=\dfrac{4\pm \sqrt{21}}{-5}$.
$\Rightarrow \dfrac{-\left( -8 \right)\pm \sqrt{{{\left( -8 \right)}^{2}}-4\left( -5 \right)\left( 1 \right)}}{2\left( -5 \right)}=\dfrac{4+\sqrt{21}}{-5},\dfrac{4-\sqrt{21}}{-5}$.
$\Rightarrow \dfrac{-\left( -8 \right)\pm \sqrt{{{\left( -8 \right)}^{2}}-4\left( -5 \right)\left( 1 \right)}}{2\left( -5 \right)}=\dfrac{-4-\sqrt{21}}{5},\dfrac{-4+\sqrt{21}}{5}$.
So, we have found the roots of the given equation $-5{{x}^{2}}-8x+1=0$ as $\dfrac{-4-\sqrt{21}}{5}$, $\dfrac{-4+\sqrt{21}}{5}$.

$\therefore $ The roots of the given equation $-5{{x}^{2}}-8x+1=0$ are $\dfrac{-4-\sqrt{21}}{5}$, $\dfrac{-4+\sqrt{21}}{5}$.

Note: Whenever we get this type of problems, we first compare the given equation with $\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ to find the required answer. We can also solve this problem without using quadratic formula by factoring the equation first and then equating each factor zero to get the required roots. We should not make calculation mistakes while solving this type of problem. Similarly, we can expect problems to find the equation $5{{x}^{2}}+7x+2=0$.