Question

How do you factor and solve ${{x}^{2}}=12x-27$?

Answer 1

Hint: We are given with a quadratic expression which we have to solve. We will start by writing the given equation in the standard form, which is, $a{{x}^{2}}+bx+c=0$. Then we will proceed solving the equation using the factorization method. As we know that, ${{x}^{2}}-(sum\_of\_zeroes)x+(product\_of\_zeroes)=0$. So, we will have two zeroes of the given equation.

Complete step by step solution:
According to the given expression we have to factorize and solve for $x$. Firstly, we will write the given expression in the standard form of a quadratic equation which is $a{{x}^{2}}+bx+c=0$.
So we have,
${{x}^{2}}=12x-27$
$\Rightarrow {{x}^{2}}-12x+27=0$-----(1)
We will look for factors of 27 such that, when the factors are added they give a sum of 12. We get,
$27=3\times 9$
$12=3+9$
We know that,
${{x}^{2}}-(sum\_of\_zeroes)x+(product\_of\_zeroes)=0$-----(2)
Using the above expression, we get,
${{x}^{2}}-12x+27=0$
$\Rightarrow {{x}^{2}}-(3+9)x+(3\times 9)=0$
$\Rightarrow {{x}^{2}}-3x-9x+27=0$
$\Rightarrow x(x-3)-9(x-3)=0$
Taking common factors aside, we get,
$\Rightarrow (x-3)(x-9)=0$
So we can have,
$x-3=0$ and $x-9=0$
We get the values of $x=3,9$
Therefore, the value of $x$ is 3 and 9.

Note:
The above question can also be solved using the discriminant, we get,
${{x}^{2}}-12x+27=0$----(1)
We have the values as $a=1,b=-12,c=27$.
We know that,
$D={{b}^{2}}-4ac$----(2)
Substituting the known values, we have, we get,
$\Rightarrow D={{(-12)}^{2}}-4(1)(27)$
$\Rightarrow D=144-108$
$\Rightarrow D=36>0$
So, we will have two real roots.
Now we will be using the quadratic formula, we have,
$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$-----(3)
Substituting the values of known terms in the equation above, we have,
$\Rightarrow x=\dfrac{-(-12)\pm \sqrt{36}}{2(1)}$
$\Rightarrow x=\dfrac{12\pm \sqrt{36}}{2}$
$\Rightarrow x=\dfrac{12\pm 6}{2}$
$\Rightarrow x=\dfrac{12+6}{2},\dfrac{12-6}{2}$
Solving further, we get the two values of $x$, we have,
$\Rightarrow x=\dfrac{18}{2},\dfrac{6}{2}$
$\Rightarrow x=9,3$
Therefore, the values of $x$ we got are 9 and 3.