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If $9x^4-12x^3+28x^2+ax+b$ is a perfect square, find the value of $a$ and $b$. This is one of my past year examination's questions, some help on it? (The answer for this problem is $a=-16$, $b=16$.)

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$9x^4-12x^3+28x^2+ax+b=(3x^2+cx+d)^2=9x^4+6cx^3+(c^2+6d)x^2+2cdx+d^2$

So $c=-2, d=4$ and $a=2cd=-16, b=d^2=16$

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You might want to try writing the polynomial as $(3x^2 + \alpha x + \beta)^2$ and solving for $\alpha$ and $\beta$.

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Even though it doesn't affect the final outcome in this problem, I think it's better long-term practice not to assume that the leading term is positive. This emphasizes that there are two square roots to a square polynomial, differing by sign, just like with integers.


Suppose $9x^4-12x^3+28x^3+ax+b=g(x)^2$ for some polynomial $g(x)$. Because the degree of $9x^4-12x^3+28x^3+ax+b$ is 4, we have $4=\deg(g(x)^2)=2\deg(g(x))$, so that $g$ is a degree 2 polynomial.

Write $g(x)=cx^2+dx+e$. Then $$9x^4-12x^3+28x^3+ax+b=(cx^2+dx+e)^2=(c^2)x^4+(2cd)x^3+(2ce+d^2)x^2+(2de)x+e^2.$$ Thus $c^2=9$, $2cd=-12$, $2ce+d^2=28$, $2de=a$, and $e^2=b$.

$c^2=9$ implies $c=3$ or $c=-3$.

Combined with $2cd=-12$, this implies that $d=2$ or $d=-2$.

Combined with $2ce+d^2=28$, this implies that $e=4$ or $e=-4$.

If $c=3$, then $d=-2$ and $e=4$. If $c=-3$, then $d=2$ and $e=-4$. Thus,

$$g(x)=\pm(3x^2-2x+4),$$

and $a=2de=-16$ and $b=e^2=16$.

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In that case:

$$ 9x^{4} - 12x^{3} + 28x^{2} + ax + b =\bracks{3x^{2} + cx + \root{b}}^{2} $$

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$$ 9x^{4} - 12x^{3} + 28x^{2} + ax =\pars{3x^{2} + cx}^{2} + 2\root{b}\bracks{3x^{2} + cx} \quad\imp\quad \dsc{\large a = 2\root{b}c} $$

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$$ 9x^{4} - 12x^{3} + 28x^{2} =\pars{3x^{2} + cx}^{2} + 6\root{b}x^{2} $$

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$$ 9x^{2} - 12x + 28=\pars{3x + c}^{2} + 6\root{b}\quad\imp\quad \dsc{\large 28=c^{2} + 6\root{b}} $$

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$$ 9x^{2} - 12x =9x^{2} + 6cx\quad\imp\quad \dsc{\large c=-2} $$

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$$ \color{#66f}{\large b}=\pars{28 - c^{2} \over 6}^{2}=\color{#66f}{\large 16} \,,\qquad \color{#66f}{\large a}=2\root{16}\pars{-2}=\color{#66f}{\large -16} $$

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