Question

Plzzz solve the problem..... ​

Answer 1

Answer:-

$\boxed{\red{\bf a = 1}}$

Given that :-

• One zero of the quadratic polynomial
• (a² + 1)x² + 56x + 2a is reciprocal to other.

To find :-

The value of a?

Solution:-

• Let the zeroes of the quadratic polynomial be α and 1/α.

We know that,

Product of zeroes = $\sf \dfrac{constant \: term}{coefficient \: of \: x^2}$

$\sf \alpha \times \frac{1}{\alpha} = \frac{2a}{a^2 +1}\\\\\implies \sf 1 = \frac{2a}{a^2 +1}\\\\\implies \sf a^2 +1=2a \\\\\implies \sf a^2 - 2a + 1 = 0$

We got a quadratic equation, solving this equation by splitting the middle term ;

$\sf \implies a^2 - a - a + 1 = 0\\\\\implies \sf a(a-1)-1(a-1)= 0 \\\\\implies \sf (a-1)(a-1)= 0\\\\\implies \sf a = 1 \: or \: a = 1$

• Hence, the required value of a is 1.

Answer 2

Answer:

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.

hope this helps you