Question

if one zero of the polynomial (a²+9)x²+ 13x+ 6a is the reciprocal of the other. Then the value of 'a' is????​

Answer 1

Answer:

• a = 3

Explanation:

Given polynomial,

$\rm \implies \: {( {a}^{2} + 9 )x}^{2} + 13x + 6a$

Whose one zero is reciprocal of another.

Let's say one zero as α

Then, another would be 1/α

We know,

$\rm \bullet \: \: product \: of \: zeros = \dfrac{ constant \: term}{coefficient \: of \: {x}^{2} }$

Then,

$\rm \implies \: \alpha \times \dfrac{1}{ \alpha } = \dfrac{6a}{ ({a}^{2} + 9) }$

Solving further,

$\rm \implies \: 1 = \dfrac{6a}{ ({a}^{2} + 9) }$

$\rm \implies \: {a}^{2} + 9 = 6a$

$\rm \implies \: {a}^{2} - 6a + 9 = 0$

Splitting middle term,

$\rm \implies \: {a}^{2} - 3a - 3a + 9 = 0$

$\rm \implies \: a(a - 3) - 3(a - 3) = 0$

$\rm \implies \: (a - 3)(a - 3) = 0$

$\rm \implies \: a = 3 \: and \: a = 3$

Hence, the value of ‘a’ is 3

Answer 2

Answer:

If one zero of the polynomial (a2 + 9)x2 + 13x + 6a is the reciprocal of the other, then the value of a is 3.

HOPE THIS WILL HELP YOU