Question

if one zero of the polynomial (A2+9)x²+13x+6a is the reciprocal of the other , find the value of a.​

Answer 1

EXPLANATION.

One zero of the polynomial.

(a² + 9)x² + 13x + 6a is reciprocal of the other.

As we know that,

Let, one root = α.

Other root is = 1/α.

Products of the zeroes of the quadratic equation.

⇒ αβ = c/a.

⇒ α x 1/α = 6a/(a² + 9).

⇒ 1 = 6a/(a² + 9).

⇒ a² + 9 = 6a.

⇒ a² - 6a + 9 = 0.

Factorizes the equation into middle term splits, we get.

⇒ a² - 3a - 3a + 9 = 0.

⇒ a(a - 3) - 3(a - 3) = 0.

⇒ (a - 3)(a - 3) = 0.

⇒ (a - 3)² = 0.

⇒ (a - 3) = 0.

⇒ a = 3.

                                                                                                                       

MORE INFORMATION.

Nature of the roots o the quadratic expression.

(1) = Real and unequal, if b² - 4ac > 0.

(2) = Rational and different, if b² - 4ac is a perfect square.

(3) = Real and equal, if b² - 4ac = 0.

(4) = If D < 0 Roots are imaginary and unequal Or complex conjugate.

Answer 2

Given :-

$\sf (a^2+9)x^{2} +13x+6a$

To Find :-

value of a

Solution :-

Other root is reciprocal of other

Product of zeroes = c/a

α × 1/α = 6a/a² + 9

α/α = 6a/a² + 9

1 = 6a/a² + 9

1(a² + 9) = 6a

a² + 9 = 6a

a² + 9 - 6a = 0

a² - 6a + 9 = 0

a² - (3a + 3a) + 9 = 0

a² - 3a - 3a + 9 = 0

a(a - 3) - 3(a + 3) = 0

(a - 3)(a - 3) = 0

Value of a

a - 3 = 0

a = 3

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