Question

$x^2+x+1+2k(x^2-x-1)=0$ is perfect square for how many values of $k$

• A. $2$
• B. $0$
• C. $1$
• D. $3$

Answer 1

The correct option is A $2$

Now, $x^2+x+1+2k{x}^2-2kx-2k=0$

$(1+2)x^2+(1-2k)x+(1-2k)=0$ ____ (1)

Now, $(ax+b{)}^2=a^2{x}^2+b^2+2abx$

Comparing with (1)

$2\sqrt{(1-2k)(1+2k)}=(1-2k)$

$4(1-4k^2)=1+4k^2-4k$

$4-16k^2=1+4k^2-4k$

$20k^2-4k-3=0$

$k=\frac{4\pm \sqrt{16+240}}{40}$

$k=\frac{1}{2}$ or $\frac{-3}{10}$

$\therefore$ two values