Question

if one zero of p(x)=(a2+4)x2+13x+4a is reciprocal of other then find the value of a

Answer 1

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$(a^2 +4)x^2 + 13x + 4a = 0$

Let the zeroes be

$\alpha, 1/\alpha$

$\alpha * 1/\alpha = c/a$

$\alpha/\alpha = 4a/(a^2 + 4)$

$1 = 4a/(a^2+4)$

$a^2 + 4 - 4a = 0$

$a^2 -4a + 4 = 0$

$a^2 -2a -2a + 4 = 0$

$a(a-2) - 2(a-2) = 0$

$(a-2)(a-2)=0$

$a=2$

Answer 2

Answer :-

Now let one of the roots be A. Then according to question the other root must be 1/A.

Now we know that

Product of roots = constant term of quadratic equation/(coefficient of x²)

Putting values we get

A x 1/A = 4a/(a² + 4) = 1 = 4a/(a² + 4)

= a² + 4 = 4a = a² - 4a + 4 = 0 ( rearranging above equation)

= a² - 2a - 2a + 4 = 0 = a(a - 2) - 2(a - 2) = 0

= (a - 2)(a - 2) = 0 = (a - 2)² =0

= a - 2 = 0 = a = 2