Question

Find the quadratic equation whose roots are given as $\dfrac{1}{2}$ and $\dfrac{-3}{4}$?
(a) $8{{x}^{2}}+2x-1=0$,
(b) $8{{x}^{2}}+2x-4=0$,
(c) $8{{x}^{2}}+2x-2=0$,
(d) $8{{x}^{2}}+2x-3=0$.

Answer 1

We start solving the problem by assuming the required quadratic equation. We then use the fact that the sum and product of the quadratic equation $a{{x}^{2}}+bx+c=0$ is $\dfrac{-b}{a}$ and $\dfrac{c}{a}$ to find the sum and product of the roots. We then substitute the obtained results in the assumed equation and make the necessary calculations to get the required quadratic equation.

Complete step-by-step solution
According to the problem, we need to find the quadratic equation whose roots are given as $\dfrac{1}{2}$ and $\dfrac{-3}{4}$.
Let us assume the quadratic equation is ${{x}^{2}}+px+q=0$ -----(1).
We know that the sum and product of the quadratic equation $a{{x}^{2}}+bx+c=0$ is $\dfrac{-b}{a}$ and $\dfrac{c}{a}$. Let us use these results for the quadratic equation in (1).
So, we get $-p=\dfrac{1}{2}-\dfrac{3}{4}$.
$\Rightarrow -p=\dfrac{2-3}{4}$.
$\Rightarrow -p=\dfrac{-1}{4}$.
$\Rightarrow p=\dfrac{1}{4}$ -------(2).
Now, we get $q=\dfrac{1}{2}\times \dfrac{-3}{4}$.
$\Rightarrow q=\dfrac{-3}{8}$ -------(3).
Let us substitute the obtained values of p and q from equations (2) and (3) in equation (1).
So, we get ${{x}^{2}}+\left( \dfrac{1}{4} \right)x+\left( \dfrac{-3}{8} \right)=0$.
$\Rightarrow {{x}^{2}}+\dfrac{x}{4}+\dfrac{-3}{8}=0$.
$\Rightarrow \dfrac{8{{x}^{2}}+2x-3}{8}=0$.
$\Rightarrow 8{{x}^{2}}+2x-3=0$.
So, we have found the required quadratic equation as $8{{x}^{2}}+2x-3=0$.
$\therefore$ The quadratic equation whose roots are given as $\dfrac{1}{2}$ and $\dfrac{-3}{4}$ is $8{{x}^{2}}+2x-3=0$. The correct option for the given problem is (d).

Note: We can also assume the quadratic equation as $p{{x}^{2}}+qx+r=0$, which will also need to be changed into the simplified form we have just assumed at the start of the problem in order to get the solution. We can also solve this problem alternatively as shown below.
We know that the quadratic equation whose roots are a and b is $\left( x-a \right)\times \left( x-b \right)=0$.
So, we get $\left( x-\dfrac{1}{2} \right)\left( x+\dfrac{3}{4} \right)=0$.
$\Rightarrow {{x}^{2}}-\dfrac{x}{2}+\dfrac{3x}{4}-\dfrac{3}{8}=0$.
$\Rightarrow \dfrac{8{{x}^{2}}-4x+6x-3}{8}=0$.
$\Rightarrow \dfrac{8{{x}^{2}}+2x-3}{8}=0$.
$\Rightarrow 8{{x}^{2}}+2x-3=0$.