Search any question & find its solution
Question:
Answered & Verified by Expert
If the equation $2 x^3+5 x^2-4 x-12=0$ has a repeated root, then the constant term of the quadratic equation whose roots are the distinct roots of the given equation is
Options:
Solution:
2804 Upvotes
Verified Answer
The correct answer is:
$-6$
Given equation $2 x^3+5 x^2-4 x-12=0$
$\begin{aligned}
& \Rightarrow 2 x^3+4 x^2+x^2+2 x-6 x-12=0 \\
& \Rightarrow\left(2 x^2+x-6\right)(x+2)=0 \\
& \Rightarrow\left(2 x^2+4 x-3 x-6\right)(x+2)=0
\end{aligned}$
$\Rightarrow(2 x-3)(x+2)(x+2)=0 \Rightarrow x=-2,-2, \frac{3}{2}$
so repeated roots $=-2$,
Now, quadratic equation
$\begin{aligned}
& x^2-\left(-2+\frac{3}{2}\right) x+(-2) \times \frac{3}{2}=0 \\
& \Rightarrow 2 x^2+x-6=0 \Rightarrow \text { Constant term }=-6
\end{aligned}$
$\begin{aligned}
& \Rightarrow 2 x^3+4 x^2+x^2+2 x-6 x-12=0 \\
& \Rightarrow\left(2 x^2+x-6\right)(x+2)=0 \\
& \Rightarrow\left(2 x^2+4 x-3 x-6\right)(x+2)=0
\end{aligned}$
$\Rightarrow(2 x-3)(x+2)(x+2)=0 \Rightarrow x=-2,-2, \frac{3}{2}$
so repeated roots $=-2$,
Now, quadratic equation
$\begin{aligned}
& x^2-\left(-2+\frac{3}{2}\right) x+(-2) \times \frac{3}{2}=0 \\
& \Rightarrow 2 x^2+x-6=0 \Rightarrow \text { Constant term }=-6
\end{aligned}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.