Question

How do you solve ${{x}^{2}}-16x+48=0$ ?

Answer 1

Hint: To solve the equation given in the above problem. First of all, we should know the degree of this equation. Degree of any equation is the highest power of the variable in the equation. After seeing the degree, you will find that the given equation is a quadratic equation and solving of this quadratic equation is done by factoring the given quadratic equation first. For that, we are going to multiply the coefficient of ${{x}^{2}}$ and the constant and then write the factors for this result of multiplication. And then arrange the factors in such a way so that addition or subtraction will give the coefficient of x. And finally, equate the factors of this quadratic equation to get the values of x.

Complete step by step answer:
The equation given above which we have to solve is as follows:
${{x}^{2}}-16x+48=0$
As you can see that the highest power of this equation is 2 so the degree of the above equation is 2 and hence, the above equation is a quadratic equation.
Now, to solve the above quadratic equation, we are going to factorize this equation by finding the factors of 48.
$\begin{align}
  & 48=12\times 4 \\
 & 48=6\times 8 \\
 & 48=48\times 1 \\
 & 48=24\times 2 \\
 & 48=16\times 3 \\
\end{align}$
If you look at the first factor then you will find that on adding this factor (12 + 4) you will get the coefficient of x which is 16 so substituting $12+4$ in place of 16 in the above equation we get:
$\begin{align}
  & {{x}^{2}}-\left( 12+4 \right)x+48=0 \\
 & \Rightarrow {{x}^{2}}-12x-4x+48=0 \\
\end{align}$
Taking x as common from first two terms in the above equation and -4 as common from the last two terms and we get,
$x\left( x-12 \right)-4\left( x-12 \right)=0$
Now, taking $\left( x-12 \right)$ as common from the above we get,
$\left( x-12 \right)\left( x-4 \right)=0$
Equating each bracket to 0 we get,
$\begin{align}
  & x-12=0 \\
 & \Rightarrow x=12 \\
 & x-4=0 \\
 & \Rightarrow x=4 \\
\end{align}$

Hence, we got two solutions of x i.e. 4 and 12 of the above equation.

Note: You can check the above solutions by substituting the values of x that we have solved in the given equation and see whether these values of x are satisfying the equation or not.
The values of x that we have solved above are:
$x=4,12$
In the below, we are just checking $x=4$ by substituting this value of x in the given quadratic equation.
$\begin{align}
  & {{x}^{2}}-16x+48=0 \\
 & \Rightarrow {{\left( 4 \right)}^{2}}-16\left( 4 \right)+48=0 \\
 & \Rightarrow 16-64+48=0 \\
 & \Rightarrow 64-64=0 \\
 & \Rightarrow 0=0 \\
\end{align}$
As you can see that L.H.S = R.H.S so the value of $x=4$ which we have calculated above is correct.
Similarly, you can check the other value of x in the same way.