Question

If $\alpha ,\beta$ are zeros of quadratic polynomial $k{x}^2+4x+4$, find the value of k such that $(\alpha +\beta {)}^2-2\alpha \beta =24.$

• A. $k=1$ or $k=\frac{11}{3}$
• B. $k=-1$ or $k=\frac{2}{3}$
• C. $k=1$ or $k=\frac{1}{3}$
• D. $k=-1$ or $k=\frac{10}{3}$

Answer 1

Answer: (B)

$k=-1$ or $k=\frac{2}{3}$

$\left(x\right)=k{x}^2+4x+4$
$\alpha +\beta =-\frac{4}{k},\alpha \beta =\frac{4}{k}$
Now, $(\alpha +\beta {)}^2-2\alpha \beta =24$
$\Rightarrow {(-\frac{4}{k})}^2-2\left(\frac{4}{k}\right)=24$
$\Rightarrow \frac{16}{k^2}-\frac{8}{k}=24$
$\Rightarrow 24k^2+8k-16=0$
$\Rightarrow 3k^2+k-2=0$
$\Rightarrow 3k^2+3k-2k-2=0$
$\Rightarrow 3k(k+1)-2(k+1)=0$
$\Rightarrow (k+1)(3k-2)=0$
$\Rightarrow k=-1$ or $k=\frac{2}{3}$