Question

How do you solve $2{{x}^{2}}-8x=-8$?

Answer 1

Hint: To solve the given quadratic equation, we first need to write it in the standard form of $a{{x}^{2}}+bx+c=0$, that is, make its RHS equal to zero. The RHS of the given equation is non-zero, and is equal to $-8$. So we need to add $8$ on both sides of the given equation to write the given equation as $2{{x}^{2}}-8x+8=0$. Then we need to simplify the equation by dividing it by $2$ so that the equation will get reduced to ${{x}^{2}}-4x+4=0$. Then, we need to substitute the values of the coefficients of ${{x}^{2}}$, $x$ and the constant terms, which are respectively $a=1$, $b=-4$ and $c=4$, into the quadratic formula $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ to finally get two solutions.

Complete step by step solution:
The equation given to us is
$\Rightarrow 2{{x}^{2}}-8x=-8$
We can see that we have a quadratic equation which is not written in the standard form, since its RHS is non-zero and is equal to $-8$. Therefore, we have to add $8$ on both sides of the above equation to get
$\begin{align}
  & \Rightarrow 2{{x}^{2}}-8x+8=-8+8 \\
 & \Rightarrow 2{{x}^{2}}-8x+8=0 \\
\end{align}$
Since $2$ is common to all the terms, we divide the above equation by $2$ to get
$\Rightarrow {{x}^{2}}-4x+4=0$
From the above equation, we can note the values of the coefficients as $a=1$, $b=-4$ and $c=4$. Substituting these into the quadratic formula, we get
$\begin{align}
  & \Rightarrow x=\dfrac{-\left( -4 \right)\pm \sqrt{{{\left( -4 \right)}^{2}}-4\left( 1 \right)\left( 4 \right)}}{2\left( 1 \right)} \\
 & \Rightarrow x=\dfrac{4\pm \sqrt{16-16}}{2} \\
 & \Rightarrow x=\dfrac{4\pm \sqrt{0}}{2} \\
 & \Rightarrow x=\dfrac{4}{2},x=\dfrac{4}{2} \\
 & \Rightarrow x=2,x=2 \\
\end{align}$

Hence, the solutions of the given equation are $x=2$ and $x=2$.

Note: We can use the given equation without using the quadratic formula too. If we can observe, the LHS of the equation obtained in the above solution, ${{x}^{2}}-4x+4=0$, is just the expansion of ${{\left( x-2 \right)}^{2}}$. So we can write the equation as ${{\left( x-2 \right)}^{2}}=0$ from which we can directly get the solutions as $x=2$ and $x=2$. Also, although the solutions are repeated, we have to write them two times since we have a two degree equation.