Question

Solve for the expression ${{x}^{2}}-2x-15=0$
(a) $x=3,5$
(b) $x=-5,3$
(c) $x=5,-3$
(d) $x=-3,-5$

Answer 1

Hint: Using mid-term splitting method to get the zeroes of the given quadratic equation. It means split the middle term of the given quadratic into two terms such that the product of them is equal to the product of the first and last term of the quadratic. Here, write $-2x$ as $-5x+3x$ and hence proceed further by taking $'x'$ as common from the first two terms and ‘3’ from the last two terms. Hence, get the roots of the quadratic.

Complete step-by-step answer:
Here, we are given quadratic equation as
${{x}^{2}}-2x-15=0.............\left( i \right)$
As we know, the factorization of any quadratic equation can be done by splitting the middle term into two terms such that multiplication of them is equal to the product of the first and last term of the quadratic.
So, by observation of the quadratic we can write $-2x$ as $-5x+3x$ and hence, multiplication of $-5x$ and $3x$ is $-15{{x}^{2}}$ which is same as the product of first and last term of the quadratic.
So, we can write equation (i) as
${{x}^{2}}-5x+3x-15=0$
Now, we can take $x$ as common from the first two terms and 3 from the last two terms. So, we can rewrite the equation as
$\begin{align}
  & x\left( x-5 \right)+3\left( x-5 \right)=0 \\
 & \Rightarrow \left( x+3 \right)\left( x-5 \right)=0..............\left( ii \right) \\
\end{align}$
Now, as we know the product of two variables can be zero, only If one of them is zero i.e. if $xy=0$ then $x=0$ or $y=0$ . Hence, as $\left( x+3 \right)$ and $\left( x-5 \right)$ are in multiplication in equation (ii), so one of them should be 0. Hence, we get equation (ii) as
$\begin{align}
  & x+3=0\text{ or }x-5=0 \\
 & x=-3\text{ or }x=5 \\
\end{align}$
Hence, roots/solutions of the given quadratic in the problem are -3,5.
So, option (c) is the correct answer.

Note: Another approach to get the roots of the given quadratic equation would be that we can use quadratic formula to get the roots, which is given as $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ for quadratic $a{{x}^{2}}+bx+c=0$ .
So, put a = 1, b = -2, c = -15 to get the zeroes/roots of the given quadratic.
One may go wrong with the mid-term splitting step. One may not be able to solve any quadratic with mid-term splitting method. We need to use the above approach for the quadratic which has irrational or imaginary roots. We can find roots by mid-term splitting if roots are rational.
Observation is the key point for a mid-term splitting approach. One needs to practice a lot of questions to visualize the roots of the equation.