Search any question & find its solution
Question:
Answered & Verified by Expert
If $\alpha$ be a root of the equation $4 x^2+2 x-1=0$, then the other root of the equation is
Options:
Solution:
2068 Upvotes
Verified Answer
The correct answer is:
$4 \alpha^3-3 \alpha$
$4 \alpha^3-3 \alpha$
$4 x^2+2 x-1=0$
$\therefore \quad \alpha+\beta=-\frac{1}{2}, \alpha \beta=-\frac{1}{4}$
Also, $4 \alpha^2+2 \alpha-1=0$ as $\alpha$ is a root and we have to prove that $\beta=4 \alpha^3-3 \alpha$.
Now, $4 \alpha^3-3 \alpha=4 \alpha^2 \cdot \alpha-3 \alpha$
$=\alpha(1-2 \alpha)-3 \alpha=-2 \alpha^2-2 \alpha$
$=-\frac{1}{2}\left[4 \alpha^2+4 \alpha\right]=-\frac{1}{2}[1-2 \alpha+4 \alpha]$
$=-\frac{1}{2}(1+2 \alpha)=-\frac{1}{2}-\alpha=\beta$
Now, $\alpha+\beta=-\frac{1}{2}$
[from Eq. (i)]
Hence, $\beta=4 \alpha^3-3 \alpha$.
$\therefore \quad \alpha+\beta=-\frac{1}{2}, \alpha \beta=-\frac{1}{4}$
Also, $4 \alpha^2+2 \alpha-1=0$ as $\alpha$ is a root and we have to prove that $\beta=4 \alpha^3-3 \alpha$.
Now, $4 \alpha^3-3 \alpha=4 \alpha^2 \cdot \alpha-3 \alpha$
$=\alpha(1-2 \alpha)-3 \alpha=-2 \alpha^2-2 \alpha$
$=-\frac{1}{2}\left[4 \alpha^2+4 \alpha\right]=-\frac{1}{2}[1-2 \alpha+4 \alpha]$
$=-\frac{1}{2}(1+2 \alpha)=-\frac{1}{2}-\alpha=\beta$
Now, $\alpha+\beta=-\frac{1}{2}$
[from Eq. (i)]
Hence, $\beta=4 \alpha^3-3 \alpha$.
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.