Question : If $x^{4}+2x^{3}+ax^{2}+bx+9$ is a perfect square, where a and b are positive real numbers, then the value of $a$ and $b$ are:
Option 1: $a=5, b=6$
Option 2: $a=6, b=7$
Option 3: $a=7, b=7$
Option 4: $a=7, b=8$
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Correct Answer: $a=6, b=7$
Solution :
Given:
$x^4+2x^3+ax^2+bx+9$
Solution:
$x^{4}+2x^{3}+ax^{2}+bx+9$
Let the equation be $[(x-\alpha)(x-\beta)]^2$
So the roots are $\alpha$, $\alpha$, $\beta$, $\beta$
$\alpha+\alpha+\beta+\beta=-2$ (Sum of roots = – coefficient of $x^3$ divided by coefficient of $x^4$)
$⇒\alpha+\beta=-1$
$\alpha \times \alpha \times \beta \times \beta=9$ (Product of roots = constant term divided by coefficient of $x^4$)
$⇒\alpha \times \beta=\pm 3$
The sum of two roots taken at a time = $a$ (sum of the roots taken two at a time = coefficient of $x^2$ divided by coefficient of $x^4$)
$a = \alpha^2+4\alpha\beta+\beta^2$
$⇒a = (\alpha+\beta)^2+2\alpha\beta$
$⇒a = 1 \pm 2 \times 3$
$⇒a = 1 + 6 = 7$ (Since, $a$ is positive $\alpha\beta=+3$)
The sum of three roots taken at a time = $-b$ (sum of the roots taken three at a time = – coefficient of $x$ divided by coefficient of $x^4$)
$-b = 2\alpha^2\beta+2\alpha\beta^2$
$⇒-b = 2\alpha\beta(\alpha+\beta)$
$⇒-b = 2\times -1\times 3$
$⇒-b = – 6$
$⇒b = 6$
Hence, the correct answer is '$a = 7$ and $b = 6$'.
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