Find the values of 'a' and 'b'

Question

$x^4-4x^3+ax^2+4x+b=0$ has two pairs of equal roots. Find the values of $a$ and $b$.

I solved like this. But I was getting a different answer than the original ($a=2, b=1$)

I used relations between roots and coefficients. I was getting a different solution.
Assumed $p,p,q,q$ as roots.

$2(p+q)=4$ $\Rightarrow$ $(p+q)=2$

$p^2+2pq+q^2=a$ $\Rightarrow$ $(p+q)^2=a$ $\Rightarrow$ $a=4$

$2pq(p+q)=-4$ $\Rightarrow$ $2pq(2)=-4$ $\Rightarrow$ $pq=1$

$(pq)^2=b$ $\Rightarrow$ $b=1$

I was getting $a=4$ and $b=1$

Point me where I made the mistake.

Answer 1

Here's another way. If $x_1, x_2$ are roots, then let $(x-x_1)(x-x_2) = x^2 + cx + d$, and we obtain

\begin{align} x^4 - 4x^3 + ax^2 + 4x + b &= (x^2 + cx + d)^2 \\ &= x^4 + 2cx^3 + (c^2+2d)x^2 + 2cdx + d^2 \end{align}

It follows that \begin{cases} 2c = -4 \\ 2cd = 4 \\ a = c^2 + 2d \\ b = d^2 \end{cases}

$\implies c = -2, \ d = -1, \ a = 2, \ b = 1 $

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Note that the roots of $x^2-2x-1$ are $1 \pm \sqrt{2}$, which are irrational, so working with them is much more cumbersome.

Edit: Using Vieta's formulas we obtain

\begin{cases} 2p + 2q = 4 \\ p^2 + 4pq + q^2 = a \\ 2p^2q + 2pq^2 = -4 \\ p^2q^2 = b \end{cases}

$\implies p+q=2, pq=-1$

$\implies a = (p+q)^2 + 2pq = 2, \ b = (pq)^2 = 1$

Answer 2

It is an easy exercise to see that given a polynomial $P $, then $P $ has a multiple root at $x_0$ iff $(x-x_0)|P,P'$. In your case, you see that $\gcd (P,P') $ must be of degree 2, and I think this will solve the problem.

Answer 3

HINT: write $$x^4-4x^3+ax^2+4x+b$$ in the form $$(x-\alpha)^2(x-\beta)^2$$ Then you will get $$x^4+x^2(-2\alpha-2\beta)+x^2(\alpha^2+4\alpha\beta+\beta^2)+x(-2\alpha^2\beta-2\alpha\beta^2)+\alpha^2\beta^2$$