Search any question & find its solution
Question:
Answered & Verified by Expert
Let \(a, b, c, d \in R\). If the equations \(2 b x^2+3 c x-d=0\) and \(2 a x^2+3 b x+4 c=0\) have a common root and \(\frac{4 b c+a d}{k\left(b^2-a c\right)}=\frac{b d+4 c^2}{4 b c+a d^{\prime}}\), then \(k=\)
Options:
Solution:
1212 Upvotes
Verified Answer
The correct answer is:
\(\frac{9}{2}\)
It is given that the given quadratic equations
\(2 b x^2+3 c x-d=0 \text { and }\)
\(2 a x^2+3 b x+4 c=0\) have a common root,
\(\left(6 b^2-6 a c\right)\left(12 c^2+3 b d\right)=(8 b c+2 a d)^2\)
Then, \(6\left(b^2-a c\right) 3\left(4 c^2+b d\right)=4(4 b c+a d)^2\)
\(\Rightarrow\)
\(\frac{4 b c+a d}{\frac{9}{2}\left(b^2-a c\right)}=\frac{b d+4 c^2}{4 b c+a d}\)
On comparing, we get \(k=\frac{9}{2}\)
Hence, option (a) is correct.
\(2 b x^2+3 c x-d=0 \text { and }\)
\(2 a x^2+3 b x+4 c=0\) have a common root,
\(\left(6 b^2-6 a c\right)\left(12 c^2+3 b d\right)=(8 b c+2 a d)^2\)
Then, \(6\left(b^2-a c\right) 3\left(4 c^2+b d\right)=4(4 b c+a d)^2\)
\(\Rightarrow\)
\(\frac{4 b c+a d}{\frac{9}{2}\left(b^2-a c\right)}=\frac{b d+4 c^2}{4 b c+a d}\)
On comparing, we get \(k=\frac{9}{2}\)
Hence, option (a) is correct.
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.