+354 The expansion of $${\left({\mathtt{2}}{\mathtt{\,-\,}}{\mathtt{px}}\right)}^{{\mathtt{6}}}$$ in ascending powers of $${\mathtt{x}}$$, as far as the $${{\mathtt{x}}}^{{\mathtt{2}}}$$, is $${\mathtt{64}}{\mathtt{\,\small\textbf+\,}}{\mathtt{Ax}}{\mathtt{\,\small\textbf+\,}}{\mathtt{135}}{\mathtt{\,\times\,}}{{\mathtt{x}}}^{{\mathtt{2}}}$$.
Given $$$$$${\mathtt{p}}$$ > $${\mathtt{0}}$$, find the value of $${\mathtt{p}}$$ and the valuse of $${\mathtt{A}}$$.
+129850 Expanding (2 - px)^6 to the first three terms and setting it equal to 64 + Ax + 135x^2, we have:
2^6 - (6)*2^5*(px) + 15*(2^4)*(px)^2 = 64 + Ax + 135x^2
64 - 192(px) + 240(px)^2 = 64 + Ax + 135x^2 ... subtract 64 from both sides
-192(px) + 240(px)^2 = Ax + 135x^2
So, equating coefficients, we have
-192p = A and 240p^2 = 135
So, solving for p^2 using the second equation, we have that p^2 = 135/240 = 9/16
And taking the positive square root, p = 3/4
And substituting for p in the first to find A, we have
-192(3/4) = A = -144
So...p = 3/4 and A = -144
+129850 Expanding (2 - px)^6 to the first three terms and setting it equal to 64 + Ax + 135x^2, we have:
2^6 - (6)*2^5*(px) + 15*(2^4)*(px)^2 = 64 + Ax + 135x^2
64 - 192(px) + 240(px)^2 = 64 + Ax + 135x^2 ... subtract 64 from both sides
-192(px) + 240(px)^2 = Ax + 135x^2
So, equating coefficients, we have
-192p = A and 240p^2 = 135
So, solving for p^2 using the second equation, we have that p^2 = 135/240 = 9/16
And taking the positive square root, p = 3/4
And substituting for p in the first to find A, we have
-192(3/4) = A = -144
So...p = 3/4 and A = -144
