Quadratic Teaser

To provide a little motivation behind the factorization.

Consider the equation as a quadratic in $a$ $x^4-2ax^2+x+a^2-a=0\implies a^2-\left(2x^2+1\right)a+\left(x^4+x\right)=0$ and solve via quadratic formula to get $a\in\left\{x^2+x,~x^2-x+1\right\}.$

Now consider $a=x^2+x$ and $a=x^2-x+1$ as quadratics in $x$ . The first one has discriminant $4a+1$ and the second one has $4a-3$ . Each must be $\ge 0$ for the solutions to be real. Thus we must have $4a+1\ge 0~\land~ 4a-3\ge 0\implies a\ge -\dfrac{1}{4}~\land~a\ge\dfrac 3 4\implies a\ge \dfrac 3 4$ so $k=3/4=0.75$ .

The given equation is $x^4-2ax^2+x+a^2-a=0$ . Now differentiating the equation we get $f(x)=4 x^3-4ax+1=0$ As the condition have been given all roots are real hence the roots of equation must also be real for $3$ critical points(maxima or minima).

We can write $f(x)$ as $f(x)= x^3-ax=-1/4$ we then find solutions of $y=x^3-ax$ and $y=-1/4$ .

For the line $y=-1/4$ to cut $y=x^3-ax$ at three points the maximum value of $y=x^3-ax$ should be less than $-1/4$ .

After differentiating $y=x^3-ax$ we get $x=(a/3)^\frac{1}{2}$
hence maximum value of $y=x^3-ax$ comes to be $x=-2(a/3)^\frac{3}{2}$
Putting this value in above condition we get $-2(a/3)^\frac{3}{2}\leq-1/4$
Simplifying this we get $a\geq3/4$ Hence $\boxed{k=3/4=0.75}$

the equation becomes $(x^2-a)^2=a-x\rightarrow x^2=a\pm\sqrt{a-x}$ $x=\pm\sqrt{a+\sqrt{a-x}}\quad or\quad\pm\sqrt{a-\sqrt{a-x}}$ the first part equals $x=\sqrt{a+\sqrt{a-\sqrt{a+....}}},y=\sqrt{a-\sqrt{a+\sqrt{a....}}}$ $x^2=a+y, y^2=a-x$ subtracting the 2, $(x-y)(x+y)=x+y\rightarrow y=x-1$ puting this, $x^2=a+x-1$ for it to have real roots, $D\geq 0\rightarrow 4a-3\geq 0$ the second part is $x=\sqrt{a-\sqrt{a-\sqrt{a-\sqrt{a-...}}}}=\sqrt{a-x}$ $x^2+x-a=0$ $D\geq 0\rightarrow 4a+1\geq 0$ $4a+1\geq 0 ∩ 4a-3\geq 0$ $a\geq .75$

The correct answer is 0.75 . $x^{4}-2ax^{2}+x+a^{2}-a = (x^{2}+x-a)(x^{2}-x+1-a) = 0$ . So, from $(x^{2}+x-a) = 0$ we get a = -0.25 and from $(x^{2}-x+1-a) = 0$ we get a = 0.75.But -0.25 will not give all real roots. Well Thanks!

nice ques :P btw isn't it from VMC FINAL-STEP part-3 @Mayank Singh