Question

How do you solve $60={{x}^{2}}-4x$?

Answer 1

Hint: The equation given in the above problem which we have to solve is a quadratic equation. It is a quadratic equation because the highest power of the variable x is 2. We are going to factorize the given quadratic equation by multiplying the coefficient of ${{x}^{2}}$ and the constant. Then we are going to find the factors of this multiplication and we will add or subtract these factors in such a fashion so that we will get the coefficient of x and then replace the coefficient of x with these factors. And then we can easily factorize. After factorization, we will equate each factor to 0.

Complete step by step solution:
The equation given above is as follows:
$60={{x}^{2}}-4x$
Rearranging the above equation we get,
$\Rightarrow {{x}^{2}}-4x-60=0$
The above equation is a quadratic equation so we are going to factorize by finding the factors of 60 first.
Factorization of 60 is as follows:
$\begin{align}
  & 60=1\times 60 \\
 & 60=2\times 30 \\
 & 60=3\times 20 \\
 & 60=4\times 15 \\
 & 60=5\times 12 \\
 & 60=6\times 10 \\
\end{align}$
Now, we are going to find the factors and addition or subtraction of them will give us the coefficient of x. As you can take the last factor which is (6 and 10) and on subtracting 10 from 6 we will get the coefficient of x so substituting $\left( 6-10 \right)$ in place of -4 in the above quadratic equation we get,
$\begin{align}
  & \Rightarrow {{x}^{2}}+\left( 6-10 \right)x-60=0 \\
 & \Rightarrow {{x}^{2}}+6x-10x-60=0 \\
\end{align}$
Taking x as common from the first two terms and -10 from the last two terms we get,
$\Rightarrow x\left( x+6 \right)-10\left( x+6 \right)=0$
Now, taking (x + 6) as common from the above equation we get,
$\Rightarrow \left( x+6 \right)\left( x-10 \right)=0$
Equating each bracket to 0 we get,
$\begin{align}
  & x+6=0 \\
 & \Rightarrow x=-6; \\
 & x-10=0 \\
 & \Rightarrow x=10 \\
\end{align}$

Hence, we got two values of x i.e. 10 and -6.

Note: We can check the values of x which we have solved above by putting these values in the parent equation and see whether these values are satisfying or not.
Checking the value of $x=10$ by substituting this value of x in the above equation we get,
$\begin{align}
  & \Rightarrow 60={{x}^{2}}-4x \\
 & \Rightarrow 60={{\left( 10 \right)}^{2}}-4\left( 10 \right) \\
 & \Rightarrow 60=100-40 \\
 & \Rightarrow 60=60 \\
\end{align}$
As you can see that L.H.S = R.H.S so the value of $x=10$ which we have calculated above is correct. Similarly, we can check the value of $x=-6$ also.