Question

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

Answer 1

First of all I don't know if the equation is equal to 0 or not assuming that it is.

Zero of a polynomial satisfies the equation so we may consider it the root of the equation.

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

Answer 2

Answer:

error

Step-by-step explanation:

the question it self had error pleasee make sure before applying questions , we can get the correct answer because it's should be (a2+9).