Question

If one zero of (a²+1) x²- 26x+ (a+3) is the reciprocal of
the other, find the positive value of a
pls help me...it's urgent​

Answer 1

Answer:

Step-by-step explanation:

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Answer 2

$\huge\boxed{Answer}$

Let the one zero is a , then the other is 1/a .

So ,

l a * 1/a = (a + 3)/(a² + 1)

1 = (a + 3)/(a² + 1)

a² + 1 = a + 3

a² - a - 2 = 0

a² - 2a + a - 2 = 0

a(a - 2) + 1(a - 2) = 0

(a + 1) (a - 2) = 0

a = -1 not possible

a = 2 ✓ ✓

Then the zeroes are 2 & 1/2 .