If one of the roots of the equation 2x^2 + (3k + 4)x + (9k^2 3k 1)

If one of the roots of the equation 2x2+(3k+4)x+(9k2–3k–1)=0 is twice the other, then which of the following can be a value of ‘k’?

Let the roots of a quadratic equation be p and q.
Since for a standard quadratic equation a\(x^2\)+bx+c=0,
Sum of roots, p+q=\(\frac{-b}{a}\)

Product of roots,pq=\(\frac{c}{a}\)

Now , in this equation 2\(x^2\)+(3k+4)x+(9\(k^2\)-3k-1)=0
If we compare it with standard quadratic equation we get,
a=2,b=3k+4,c=9\(k^2\)-3k-1

Also it is given, one root is twice of the other
so let roots of this equation be p and 2p

Sum of roots, p+2p=\(\frac{-(3k+4)}{2}\) ............(1)

Product of root, p*2p=\(\frac{9k^2-3k-1}{2}\) ............(2)


Solving equation (1)
p+2p=\(\frac{-(3k+4)}{2}\)

3p=\(\frac{-(3k+4)}{2}\)

p=\(\frac{-(3k+4)}{6}\)

Squaring both sides
\(p^2\)=\(\frac{(3k+4)^2}{36}\)

\(p^2\)=\(\frac{9k^2+16+24k}{36 }\) ...........(3)


Now, solving equation (2)
\(p*2p\) =\(\frac{9k^2-3k-1}{2}\)

\(2p^2\) =\(\frac{9k^2-3k-1}{2 } \)

\(p^2\) =\(\frac{9k^2-3k-1}{4 } \) ...........(4)

Equating value of \(p^2\) from equation(3) and equation(4)

\(\frac{9k^2+16+24k}{36}\) = \(\frac{9k^2-3k-1}{4}\)

\(\frac{9k^2+16+24k}{9}\) = \(9k^2-3k-1\)

\(9k^2+16+24k\) = \(81k^2-27k-9\)
\(72k^2-51k-25\) =0
\(72k^2-75k+24k-25\) =0
3k (24k - 25)+1(24k - 25) =0
(3k + 1) (24k - 25) =0

k= \(\frac{-1}{3}\) and k=\(\frac{25}{24}\)

Among the options given, k=\(\frac{-1}{3}\)

Hence, C