I have to solve this question

Find all real values of $a$ such that $$x^4-2ax^2+x+a^2-a=0$$

I have been trying this question for many days but was unable to solve it.

I think we can convert this polynomial into a perfect square of some other polynomial or we can convert it into a quadratic polynomial and the do $D\geq 0$ so we can find a range bound in a and then can give the answer may be.

Or another method is we can try to factorise it by getting a value of x and then we can maybe further solve.

I have tried all methods but was unable to crack the problem. I think the quadratic method is best but I am unable to get any result from that.

So please help me by giving me approach to this question and also share the way why you used this particular method only and not the other ones and what approach one should develop while doing these questions.



Your equation$$x^4-2ax^2+x+a^2-a=0$$is a quadratic equation in $a$:$$a^2-(1+2x^2)a+x^4+x=0$$and therefore you can apply the quadratic formula. Note that$$(1+2x^2)^2-4(x^2+x)=4x^2-4x+1=(2x-1)^2.$$Therefore, the solutions are $a=x^2-x+1$ and $a=x^2+x$.

  • $\begingroup$ Thanks for your help but can you share how the thought came that it's a quadratic in a and can be easily factorised $\endgroup$ – Aryan 24k Jun 12 at 7:49
  • 3
    $\begingroup$ I looked at it and I saw that the greatest exponent in the terms with an $a$ in it is $2$. So, it is a quadratic equation in $a$. And then I simply applied the quadratic formula. $\endgroup$ – José Carlos Santos Jun 12 at 7:51
  • $\begingroup$ Is there any other way to do it if one can't see that it is factorisable $\endgroup$ – Aryan 24k Jun 12 at 8:01
  • 1
    $\begingroup$ I did not see that that it is factorisable. What I did was to solve the quadratic equation$$a^2-(1+2x^2)a+x^4+x=0.$$ $\endgroup$ – José Carlos Santos Jun 12 at 8:08
  • $\begingroup$ -Ok thanks is the answer a is greater than equal to 3/4 $\endgroup$ – Aryan 24k Jun 12 at 8:11

You can factorise your expression into two second order polynomials $$x^4-2ax^2+x+a^2-a$$$$=(-x^2-x+a)(-x^2+x-1+a)=0$$ and you can use the quadratic equation as needed.

  • $\begingroup$ Is there any other way to do it if one can't see that it is factorisable $\endgroup$ – Aryan 24k Jun 12 at 8:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.