Question

Complete set of real values of '$a$' for which the equation $x^4-2a{x}^2+x+a^2-a=0$ has all its roots real:

• A. $[\frac{3}{4},\infty )$
• B. $[1,\infty )$
• C. $[2,\infty )$
• D. $[0,\infty )$

Answer 1

The correct option is A $[\frac{3}{4},\infty )$

Given, the equation $x^4-2ax+x+a^2-a=0$

Considering it a quadratic equation in $a$,

$a^2-a(1+2x^2)+x^4+x=0$

Solving the equation we get,

$a=\frac{-[-(1+2x^2\left)\right]\pm \sqrt{{[-(1+2x\left)\right]}^2-4(x^4+x)}}{2}=\frac{(1+2x^2)\pm \sqrt{1+4x^2+4x^4-4x^4-4x}}{2}=\frac{(1+2x^2)\pm \sqrt{4x^2-4x+1}}{2}=\frac{(1+2x^2)\pm \sqrt{{(2x-1)}^2}}{2}=\frac{(1+2x^2)\pm (2x-1)}{2}$

$\therefore a=\frac{1+2x^2+2x-1}{2},\frac{1+2x^2-2x+1}{2}\Rightarrow a=x^2+x,x^2-x+1$

Thus, we get a factorisation,

$(x^2+x-a)(x^2-x+1-a)=0$

For, $x^2+x-a=0\Rightarrow x=\frac{-1\pm \sqrt{1+4a}}{2}$

For, the roots to be real $1+4a\ge 0\Rightarrow a\ge -\frac{1}{4}$

So, $a$ has the roots in $[-\frac{1}{4},\infty )$

For, $x^2-x+1-a=0\Rightarrow x=\frac{-1\pm \sqrt{4a-3}}{2}$

For, the roots to be real, $4a-3\ge 0\Rightarrow a\ge \frac{3}{4}$

So, $a$ has the roots in $[\frac{3}{4},\infty )$

We see that the common interval is $[\frac{3}{4},\infty )$

Hence, the complete set of real values of $a$ for which the given equation has all its real roots is $[\frac{3}{4},\infty ).$