Question

How do you factor and solve ${{x}^{2}}+4=5x$?

Answer 1

Hint: The given equation in the above question is a quadratic equation which must first be written in the standard form of $a{{x}^{2}}+bx+c=0$. For this, we have to make the RHS of the given equation equal to zero by subtracting \[5x\] from both sides of the equation ${{x}^{2}}+4=5x$. After subtracting we will obtain ${{x}^{2}}-5x+4=0$. Then for factoring, we have to use the middle term splitting method so as to split the middle term $-5x$ into two terms so that we will get four terms on the LHS. Then we have to use the method of factor by grouping to factor the LHS. Using the zero product rule, we will finally get the solutions of the given equation.

Complete step by step solution:
The equation given in the above question is
\[\Rightarrow {{x}^{2}}+4=5x\]
Subtracting \[5x\] from both the sides, we get
$\Rightarrow {{x}^{2}}-5x+4=0$
Using the middle term splitting method, we split the middle term in the LHS of the above equation as
$\Rightarrow {{x}^{2}}-4x-x+4=0$
Now, using the factor by grouping method, we group the first two and the last two terms to get
\[\Rightarrow \left( {{x}^{2}}-4 \right)+\left( -x+4 \right)=0\]
Now, we can take $x$ and $-1$ common from the first and the second groups to get
\[\Rightarrow x\left( x-4 \right)-1\left( x-4 \right)=0\]
Taking \[\left( x-4 \right)\] common, we get
\[\Rightarrow \left( x-4 \right)\left( x-1 \right)=0\]
Using the zero product rule, we equate each factor to zero to get
$\begin{align}
  & \Rightarrow x-4=0 \\
 & \Rightarrow x=4 \\
\end{align}$
And
$\begin{align}
  & \Rightarrow x-1=0 \\
 & \Rightarrow x=1 \\
\end{align}$
Hence, the solutions of the given equation are $x=1$ and $x=4$.

Note:
Do not take the RHS of the given equation to be zero. It is a common mistake of considering the RHS of a quadratic equation equal to zero, which is non zero in the given equation. Also, while taking the factors common from the group of the terms, the sign of the common factor is decided by the sign of the first term of that group.