Question

How do you solve ${{x}^{2}}+2x-8=0$?

Answer 1

Hint: We will apply the middle term splitting method as it is easily applicable here i.e., we will split the middle term, here the middle term is $2x$. After that, we will find the factors of the given equation i.e., $\Rightarrow {{x}^{2}}+2x-8=0$. So, let’s see.

Complete Step by Step Solution:
The given equation is ${{x}^{2}}+2x-8=0$
Now, multiply the coefficient of ${{x}^{2}}$ with coefficient of ${{x}^{0}}$ , we get $1\times \left( -8 \right)=-8$
Positive factors of $-8$ are 1, 2, 4 and 8
As the coefficient of ${{x}^{0}}$ is $-8$, therefore we will subtract the factors of $-8$ in such a way that the difference will become the coefficient of $x$
Now, we will subtract 2 from 4
$\Rightarrow {{x}^{2}}+\left( 4-2 \right)x-8=0$
$\Rightarrow {{x}^{2}}+4x-2x-8=0$
Now, we will take x common from the first two terms and $-2$ from the last two terms i.e.
$\Rightarrow x\left( x+4 \right)-2\left( x+4 \right)=0$
$\Rightarrow \left( x-2 \right)\left( x+4 \right)=0$
Either $\left( x-2 \right)=0$ or $\left( x+4 \right)=0$

So, we got two values of $x$ i.e., $x=2,-4$

Additional Information:
Sometimes the product of the coefficient of ${{x}^{2}}$ and coefficient of ${{x}^{0}}$ is too big such that making factors is a bit harder for them. So, what we do is, either we just try to take something common from it or we try to divide both sides of the equation by a suitable number. For example: In the equation $10{{x}^{2}}+30x+20=0$, we will divide both the sides by $10$ because $10$ is the greatest common number in the equation. After that, the remaining equation is ${{x}^{2}}+3x+2=0$, which we will solve similarly as we did in the above solution.

Note:
There is an alternative method to solve the given equation in which we use Quadratic Formula i.e. $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ where $a$ is the coefficient of ${{x}^{2}}$, $b$ is coefficient of $x$ and $c$ is coefficient of ${{x}^{0}}$
Quadratic Formula is $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
The given equation is ${{x}^{2}}+2x-8=0$
Here $a=1$ , $b=2$ and $c=-8$
Hence, $x=\dfrac{-2\pm \sqrt{{{2}^{2}}-4\left( 1 \right)\left( -8 \right)}}{2\left( 1 \right)}$
$\Rightarrow x=\dfrac{-2\pm \sqrt{4+32}}{2}$
On further simplification,
$\Rightarrow x=\dfrac{-2\pm \sqrt{36}}{2}$
Now, one value of $x=\dfrac{-2+6}{2}$ and another value of $x=\dfrac{-2-6}{2}$ i.e.
$\Rightarrow x=\dfrac{4}{2}$ and $x=\dfrac{-8}{2}$
$\Rightarrow x=2,-4$