Question

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

Answer 1

If one of the zero is reciprocal of the other, 
let one zero be α
other zero will be 1/α

Product of roots = c/a
⇒ α × 1/α = 6a/(a²+9)
⇒ 1 = 6a/(a²+9)
⇒ 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.

Answer 2

Answer :-

→ a = 3 .

Step-by-step explanation :-

Let one zero of the following polynomial be α .

Then, the other zero is 1/α .

$\therefore$ 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 .