Question

How do you solve $5{{x}^{2}}-40x+80=0$?

Answer 1

Hint: We first divide both sides of the equation $5{{x}^{2}}-40x+80=0$ with 5. We then form a square for the left side of the new equation. Then we take the square root on both sides of the equation. From that we add 4 to the both sides to find the value of $x$ for $5{{x}^{2}}-40x+80=0$.

Complete step by step solution:
We need to find the solution of the given equation $5{{x}^{2}}-40x+80=0$.
We first divide both sides of the equation $5{{x}^{2}}-40x+80=0$ with 5.
$\begin{align}
  & \dfrac{5{{x}^{2}}-40x+80}{5}=0 \\
 & \Rightarrow {{x}^{2}}-8x+16=0 \\
\end{align}$
Now we use the identity ${{a}^{2}}-2ab+{{b}^{2}}={{\left( a-b \right)}^{2}}$ for ${{x}^{2}}-8x+16=0$.
$\begin{align}
  & {{x}^{2}}-8x+16=0 \\
 & \Rightarrow {{\left( x \right)}^{2}}-2\times x\times 4+{{\left( 4 \right)}^{2}}=0 \\
 & \Rightarrow {{\left( x-4 \right)}^{2}}=0 \\
\end{align}$
We interchanged the numbers for $a=x,b=4$.
Now we have a quadratic equation ${{\left( x-4 \right)}^{2}}=0$.
We need to find the solution of the given equation ${{\left( x-4 \right)}^{2}}=0$.
We take square root on both sides of the equation. As the equation is a quadratic one, the number of roots will be 2 and they are equal in value but opposite in sign.
$\begin{align}
  & \sqrt{{{\left( x-4 \right)}^{2}}}=0 \\
 & \Rightarrow \left( x-4 \right)=0 \\
\end{align}$
Now we add 4 to the both sides of the equation $\left( x-4 \right)=0$ to get value for variable $x=4$.
The given quadratic equation has two solutions and they are both the same where $x=4$.

Note: We can also apply the quadratic equation formula to solve the equation $5{{x}^{2}}-40x+80=0$.
We know for a general equation of quadratic $a{{x}^{2}}+bx+c=0$, the value of the roots of x will be $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$.
In the given equation we have $5{{x}^{2}}-40x+80=0$. The values of a, b, c is $5,-40,80$ respectively.
$x=\dfrac{-\left( -40 \right)\pm \sqrt{{{\left( -40 \right)}^{2}}-4\times 5\times 80}}{2\times 5}=\dfrac{40\pm \sqrt{0}}{10}=\dfrac{40}{10}=4$.
The given quadratic equation has two solutions and they are $x=4$.