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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.

The expansion of (2 px)^6 in ascending powers of x, as far as the term in x², is 64 + Ax-example-1
User John Rajan
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1 Answer

26 votes
26 votes

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}

User Yeong
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