Question

value of 'k' so that the equation ${(x^2-2x)}^2-3(x^2-2x)+(k+2)=0$ has two real solutions

• A. $(-\infty ,-6)$
• B. $\frac{1}{4}$
• C. $(-\infty ,-6)\cup \left\{\frac{1}{4}\right\}$
• D. none of these

Answer 1

The correct option is B $\frac{1}{4}$

${\left(x^2-2x\right)}^2-3\left(x^2-2x\right)+k+2=0$---------------$\left(1\right)$

Let $y=x^2-2x$---------------------$\left(2\right)$

then$,$ above $e{q}^n\left(1\right)$ will be

$y^2-3y+\left(k+2\right)=0$-----------------$\left(3\right)$

$\because e{q}^n\left(1\right),$ have two $(=2)$ real solutions for this $e{q}^n\left(2\right)$ must have unique value of $y.$

thus$,$ $e{q}^n\left(3\right)$ must be a quadratic with two equal roots

So$,$ $y^2-3y+\left(k+2\right)=0$ have equal solutions

$\therefore D=b^2-4ac=0$

$\Rightarrow {\left(-3\right)}^2-4\times 1\times \left(k+2\right)=0$

$\Rightarrow 9-4\left(k+2\right)=0$

$\Rightarrow 9-4k-8=0$

$\Rightarrow 4k=1$

$\Rightarrow k=\frac{1}{4}$

Hence,

option $\left(B\right)$ iscorrect answer.