Question

If one zero of the polynomial $3x^2-8x+2k+1$ is seven times the other, then the value of $k$ is ___.

• A. $-\frac{1}{2}$
• B. $\frac{1}{2}$
• C. $\frac{2}{3}$
• D. $-\frac{2}{3}$

Answer 1

Answer: (A), (C)

$\frac{2}{3}$

Let one zero be ${}^{\prime }{a}^{\prime }$, so other zero will be ${}^{\prime }7a^{\prime }$.
Given, f(x) = $3x^2-8x+2k+1$
$f\left(a\right)=0$
$\Rightarrow 3a^2-8a+2k+1=0\dots \left(i\right)$

$f\left(7a\right)=0$
$147a^2-56a+2k+1=0\dots \left(ii\right)$

Subtract Eq(i) from Eq (ii),
$147a^2-56a+2k+1=0\underset{–}{{}_{-}3a^2\mp 8a\pm 2k\pm 1=0}144a^2-48a=0$

$\Rightarrow 48a(3a-1)=0$
$\therefore a=0$ and $a=\frac{1}{3}$

If we put $a=0$, then
$f\left(a\right)=3\times 0^2-8\times 0+2k+1$
$=2k+1$
$\Rightarrow 2k+1=0$
$\Rightarrow k=-\frac{1}{2}$
If $a=\frac{1}{3}$
Now$f\left(\frac{1}{3}\right)=0$

$3\times \frac{1}{3^2}-\frac{8}{3}+2k+1=0\frac{1}{3}-\frac{8}{3}+1+2k=0-\frac{7}{3}+1+2k=0-\frac{4}{3}+2k=02k=\frac{4}{3}\therefore k=\frac{2}{3}$