Given that $2$ is a root of the equation $3 x^{2} - p (x + 1) = 0$ and that the equation
$p x^{2} - q x + 9 = 0$ has equal roots, find the values of $p$ and $q .$

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Solution

Step 1: Find the values of $p$
Given $2$ is root of equation $3 x^{2} - p (x + 1) = 0$
Put $x = 2$ in the equation $3 x^{2} - p (x + 1) = 0$
$\Rightarrow p = \frac{12}{3} = 4$

Step 2 : Find the value of $q$ .
Given second equation is $p x^{2} - q x + 9 = 0$
$\Rightarrow 4 x^{2} - q x + 9 = 0$
Comparing $4 x^{2} - q x + 9 = 0$ with $a x^{2} + b x + c = 0$
$a = 4, b = - q$ and $c = 9$ .
Since roots are equal, $b^{2} - 4 a c = 0$
$\Rightarrow (- q)^{2} - 4 \times 4 \times 9 = 0$
$\Rightarrow q^{2} - 144 = 0$
$\Rightarrow q^{2} = 144$
$\Rightarrow q = \pm 12$
Hence, $p = 4$ and $q = \pm 12$ .

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