Question

13. For what values of x, the numbers (x+ 9), (x- 6) and 4 are in G.P.?​

Answer 1

Answer:

Step-by-step explanation:

we must know that if three numbers namely a, b, and c are in G.P, then

$ac=b^{2}$

using the same concept

4 x (x+9) = $(x-6)^{2}$

4x + 36 = $x^{2}$ + 36 - 12x

therefore

$x^{2}$ - 16x = 0

x(x-16) = 0

therefore, either x = 0 or 16

CHEERS MATE< PLEASE MARK BRAINLIEST