If one zero of polynomial (a^2+a)x^2+13x+6a is reciprocal of other ,find a.

I don't understand please do it again.....♥

well, the product of the roots is

(6a)/(a^2+a) = 1
6a = a^2+a
a^2-5a = 0
a(a-5) = 0
a=5

30x^2+13x+30
has roots (-13±√3431 i)/60

which are reciprocals

To find the value of 'a' for which one zero of the polynomial is the reciprocal of the other, we can follow these steps:

1. Remember that the reciprocal of a number 'x' is written as 1/x.

2. Let's assume that one of the zeros is 'k'. Therefore, the other zero will be its reciprocal, which is 1/k.

3. The polynomial given is (a^2 + a)x^2 + 13x + 6a.

4. To find the zeros, we can set this polynomial equal to zero and solve for 'x'. So, (a^2 + a)x^2 + 13x + 6a = 0.

5. Now, we know that the sum of the zeros of a quadratic equation is equal to -b/a, where 'b' is the coefficient of 'x' and 'a' is the coefficient of x^2.

6. In this case, the sum of the zeros is equal to -13/(a^2 + a).

7. Since one zero is 'k' and the other is 1/k, their sum can be written as k + 1/k.

8. Equating the above expressions, we have k + 1/k = -13/(a^2 + a).

9. Eliminating the fraction, multiply both sides of the equation by (a^2 + a) and simplify. This gives us k(a^2 + a) + 1 = -13.

10. Expanding the equation, we get a^2k + ak + 1 = -13.

11. Rearranging the terms, we have a^2k + ak + 14 = 0.

Now we have a quadratic equation in terms of 'a' and 'k'. To find the value of 'a', we need more information or constraints about 'k'. If you have any additional information or constraints about 'k', please provide it so that we can proceed further in finding 'a'.