Question

Complete set of real values for the equation9^x+a.3^x+1=0 has

Answer 1

Answer:

The roots of the quadratic equation9/x²-27=25/x²-11​

9/x²-27 = 25/x²-11

We cross multiply each other

25×(x²-27)= 9(x²-11)

25x²- 675 = 9x²- 99

16x²- 576 = 0

x²-36 =0

( By taking 16 as common )

x²=36

x =√36

x =±6

Answer 2

Given: 9^x+a.3^x+1=0

To Find: set of real values for the equation

Solution:

The equation is in the form ax²+bx+c=0, where a, b, and c are coefficients. Then the coefficients in the formula are (-b±√(b²-4ac))/(2a).

The roots of the quadratic equation 9^x+a.3^x+1=0

We multiply simply,

25×(x²-27)= 9(x²-11)/a

25x²- 675 = 9x²- 99/a

16x²- 576a = 0

x²-36 =0

x²=36a

x =√36a

x =±6a

Therefore, the values of x are 6a and -6a.

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