Question

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

Answer 1

Answer :-

→ a = 3 .

Step-by-step explanation :-

Let one zero of the following polynomial be α .

Then, the other zero is 1/α .

∴ product of zeros = ( α × 1/α ) = 1 .

But, product of zeros = ( constant term )/( coefficient of x² ) = 6a/(a² + 9 ) .

$\therefore$ 6a/(a² + 9) = 1 .

⇒ a² + 9 = 6a .

⇒ a² + 9 - 6a = 0 .

⇒ a² + 3² - 2(a)(3) = 0 .

⇒ ( a - 3 )² = 0 .

⇒ a - 3 = 0 .

$\therefore$ a = 3 .

Hence, a = 3 .

Answer 2

Answer :-

$\textsf{If one of the zero is reciprocal of the other,}$

$\textsf{let one zero be a.}$

$\sf{Other \: zero \: will \: be \: \dfrac{1}{a}.}$

Product of roots = c/a

⇒ α × 1/α = 6a/(a² + 9)

⇒ 1 = 6a/(a² + 9)

⇒ a² + 9 = 6a

⇒ a² - 6a + 9 = 0

⇒ a² + 9 = 6a

⇒ a² - 6a + 9 = 0

⇒ a² - 3a - 3a + 9 = 0

⇒ a(a - 3) - 3(a - 3) = 0

⇒ (a - 3)(a - 3) = 0

⇒ (a - 3)² = 0

⇒ a = 3

Value of a is 3.