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 $b = 6$'.