Question

How do you solve ${{x}^{2}}-5x=-20$ equation using quadratic formula?

Answer 1

Hint: The quadratic formula is given by $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$, where a is the coefficient of ${{x}^{2}}$, b is coefficient of x and c is the constant in the quadratic equation written in the standard form of $a{{x}^{2}}+bx+c=0$. Therefore, before applying the quadratic formula for solving the given equation, we must write it in the standard form, that is, make the RHS of the given equation equal to zero. For this, in the given quadratic equation ${{x}^{2}}-5x=-20$ we need to add $20$ both the sides so that the equation will become ${{x}^{2}}-5x+20=0$. Then on substituting $a=1$, $b=-5$ and $c=20$ into the quadratic formula $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$, we will obtain the two solutions of the given quadratic equation.

Complete step by step solution:
The equation given in the above question is written as
$\Rightarrow {{x}^{2}}-5x=-20$
We can see that the above equation is not written in the standard form, that is, its RHS is not equal to zero. Therefore, we add $20$ on both the sides of the above equation to get
\[\begin{align}
  & \Rightarrow {{x}^{2}}-5x+20=-20+20 \\
 & \Rightarrow {{x}^{2}}-5x+20=0.......\left( i \right) \\
\end{align}\]
So now we obtained the given equation in the standard form so that we can apply the quadratic formula which is given by
$\Rightarrow x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
From the equation (i) we can note down the values of the coefficients a, b and c respectively as
$\begin{align}
  & \Rightarrow a=1 \\
 & \Rightarrow b=-5 \\
 & \Rightarrow c=20 \\
\end{align}$
Substituting the above values of the coefficients into the quadratic formula, we get
\[\begin{align}
  & \Rightarrow x=\dfrac{-\left( -5 \right)\pm \sqrt{{{\left( -5 \right)}^{2}}-4\left( 1 \right)\left( 20 \right)}}{2\left( 1 \right)} \\
 & \Rightarrow x=\dfrac{5\pm \sqrt{25-80}}{2} \\
 & \Rightarrow x=\dfrac{5\pm \sqrt{-55}}{2} \\
 & \Rightarrow x=\dfrac{5\pm \sqrt{55}\sqrt{-1}}{2} \\
\end{align}\]
We know that \[\sqrt{-1}=i\]. On substituting this above, we get
$\Rightarrow x=\dfrac{5\pm \sqrt{55}i}{2}$

Hence, we have obtained the two solutions of the given equation as \[x=\dfrac{5+\sqrt{55}i}{2}\] and \[x=\dfrac{5-\sqrt{55}i}{2}\].

Note: Do not forget to write the given quadratic in the standard form by making the RHS equal to zero. This is because the quadratic formula $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ is defined only for the quadratic equations written in the standard form of $a{{x}^{2}}+bx+c=0$. Take care of the signs of the coefficients while substituting them into the quadratic formula.