Question

How do you solve ${{x}^{2}}=4x+12$?

Answer 1

First take all the terms to the L.H.S. Use the middle term split method to factorize ${{x}^{2}}=4x+12$ and write as a product of two terms given as $\left( x-a \right)\left( x-b \right)$, where ‘a’ and ‘b’ are called zeroes of the polynomial. Now, substitute each term equal to 0 and find the values of x to get the answer.

Complete step by step solution:
Here, we have been provided with the quadratic equation: ${{x}^{2}}=4x+12$ and we are asked to solve it. That means we have to find the values of x.
First let us take all the terms to the L.H.S, so we get,
${{x}^{2}}-4x-12=0$
Now, let us apply the middle term split method to factorize the given equation first. Let us assume $f\left( x \right)={{x}^{2}}-4x-12$.
$\because f\left( x \right)={{x}^{2}}-4x-12$
According to the middle term split method we have to break -4x into two parts such that their sum is -4x and the product is equal to the product of the constant term -12 and ${{x}^{2}}$, i.e. $-12{{x}^{2}}$. So, breaking -4x into -6x and 2x, we have,
(i) $\left( -6x \right)+\left( 2x \right)=-4x$
(ii) $\left( -6x \right)\times \left( 2x \right)=-12{{x}^{2}}$
Therefore, both the conditions of the middle term split method are satisfied, so we can write the given expression as: -

& \Rightarrow f\left( x \right)={{x}^{2}}-6x+2x-12 \\\ & \Rightarrow f\left( x \right)=x\left( x-6 \right)+2\left( x-6 \right) \\\ \end{aligned}$$ Taking $$\left( x-6 \right)$$ common in the R.H.S., we get, $$\begin{aligned} & \Rightarrow f\left( x \right)=\left( x-6 \right)\left( x+2 \right) \\\ & \Rightarrow f\left( x \right)=0 \\\ & \Rightarrow \left( x-6 \right)\left( x+2 \right)=0 \\\ \end{aligned}$$ Substituting each term equal to 0, we get, $\Rightarrow (x – 6) = 0 or (x + 2) = 0$ $\Rightarrow x = 6 or x = -2$ **Hence, the solutions of the given equations are: - x = -2 or x = 6.** **Note:** One may note that here we have applied the middle term split method to get the values of x. You can also apply the discriminant method to get the answer. In that conditions assume the coefficient of $${{x}^{2}}$$ as ‘a’, the coefficient of x as b and the constant term as c, and apply the formula: - $$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$$ to solve for the values of x. There can be a third method also, known as completing the square method, to solve the question. Note that the discriminant formula is obtained from completing the square method.