Question

pl answer the 7th question there

Answer 1

Let's find a general condition here, For the given quadratic equation, let one root be the reciprocal of the other, namely
$\alpha \: and \: \frac{1}{ \alpha }$
Quadratic equation:
${ax}^{2} + bx + c = 0$
Now since,
Product of roots=
$\frac{ c}{a}$
Therefore,
$\alpha \times \frac{1}{ \alpha } = \frac{c}{a} \\ a = c$
Therefore when one root is reciprocal of the other, the coefficient of x^2 equals the constant.

Now we can use this to solve the problem, In the given equation,
${a}^{2} + 9 = 6a \\ {a}^{2} - 6a + 9 = 0 \\ ({a - 3})^{2} = 0 \\ a = 3$
Therefore the value of
$a$
here is 3.

Remember the first general condition I have shown here that is helpful for solving many such questions.