Question

The expansion of (2 px)^6 in ascending powers of x, as far as the term in x², is

64 + Ax + 135x²
Given that p > 0, find the value of p and the value of A.

Answer 1

Answer:
`x=\frac{-A+\left(A^2-34560\right)^{(1)/(2)}}{270},\:x=\frac{-A-\left(A^2-34560\right)^{(1)/(2)}}{270}`

Explanation:

64+Ax+135x^2

135x^2+Ax+64=0

`x_(1,\:2)=(-A\pm √(A^2-4\cdot \:135\cdot \:64))/(2\cdot \:135)`

`x_(1,\:2)=\frac{-A\pm \left(A^2-34560\right)^{(1)/(2)}}{2\cdot \:135}`

`x_1=\frac{-A+\left(A^2-34560\right)^{(1)/(2)}}{2\cdot \:135},\:x_2=\frac{-A-\left(A^2-34560\right)^{(1)/(2)}}{2\cdot \:135}`

`x=\frac{-A+\left(A^2-34560\right)^{(1)/(2)}}{270},\:x=\frac{-A-\left(A^2-34560\right)^{(1)/(2)}}{270}`