Question

How do you solve ${{x}^{2}}+18=9x$ ?

Answer 1

Hint: To solve the above equation, first of all, we make all the terms in x and the constant terms on one side of the equation and 0 on the other side of the equation which can be achieved by subtracting 9x on both the sides of the given equation. Then we will solve this quadratic equation, by multiplying the coefficient of ${{x}^{2}}$ and the constant terms and factorize this multiplication is such a way so that on addition and subtraction of these factors will give you the number which is coefficient of x, Then take the common terms and hence, will find the value of x.

Complete step by step answer:
The equation given in the above problem is as follows:
${{x}^{2}}+18=9x$
The above equation is a quadratic equation because the degree of this equation is 2. Now, subtracting 9x on both the sides of the above equation we get,
$\begin{align}
  & \Rightarrow {{x}^{2}}+18-9x=9x-9x \\
 & \Rightarrow {{x}^{2}}-9x+18=0 \\
\end{align}$
Multiplying the coefficient of ${{x}^{2}}$ ( i.e. 1) and the constant term (i.e. 18) we get,
$\Rightarrow 1\left( 18 \right)=18$
Now, taking the factors of 18 we get,
$\begin{align}
  & 18=1\times 18 \\
 & 18=2\times 9 \\
 & 18=3\times 6 \\
\end{align}$
If you add the last factors of 18 (i.e. 3 and 6) then you will get 9 and which is the coefficient of x with a negative sign so substituting $\left( 6+3 \right)$ in place of 9 in the above quadratic equation we get,
$\Rightarrow {{x}^{2}}-\left( 6+3 \right)x+18=0$
Multiplying x by 6 and 3 we get,
$\Rightarrow {{x}^{2}}-6x-3x+18=0$
Taking x as common from the first two terms and -3 from the last two terms we get,
$\Rightarrow x\left( x-6 \right)-3\left( x-6 \right)=0$
As you can see that $\left( x-6 \right)$ is common in the above equation so we can take $\left( x-6 \right)$ as common from the above equation and we get,
$\Rightarrow \left( x-6 \right)\left( x-3 \right)=0$
Equating each of the above brackets to 0 we get,
$\begin{align}
  & \Rightarrow x-6=0 \\
 & \Rightarrow x=6 \\
 & \Rightarrow x-3=0 \\
 & \Rightarrow x=3 \\
\end{align}$

Hence, the solutions of the above equation are as follows:
$x=6,3$

Note: We can check the solutions which we have solved above by substituting these values of x in the above equation and see whether these values of x are satisfying or not.
Substituting the value of $x=6$ in the above equation we get,
$\begin{align}
  & \Rightarrow {{\left( 6 \right)}^{2}}+18=9\left( 6 \right) \\
 & \Rightarrow 36+18=54 \\
 & \Rightarrow 54=54 \\
\end{align}$
As you can see that L.H.S = R.H.S so $x=6$ is satisfying the given equation. Similarly, you can check the other value of x too.