Question

y = px^2 + q*(squareroot)x

estimate the value of x if:

2y = 43*(squareroot)x

Answer 1

\[y=px^2+q\sqrt{x}\]\[2y=43\sqrt{x} \Rightarrow y=\frac{43}{2}\sqrt{x}\]We have "y=" twice, so they equal each other:\[px^2+q\sqrt{x}=\frac{43}{2}\sqrt{x}\]Bring everything to one side, and take out \(\sqrt{x}\) as being common:\[\sqrt{x}\left(px^{\frac{3}{2}}+q-\frac{43}{2}\right)=0\]
So \(\sqrt{x}\) = 0, or \(px^{\frac{3}{2}}+q-\frac{43}{2}=0\)

Answer 2

so if i tell you that p = 2 and q = 3, could you solve it?

Answer 3

Well when you get to the stage\[px^{\frac{3}{2}}+q-\frac{43}{2}=0\]you can get x on its own:\[px^{\frac{3}{2}}=\frac{43}{2}-q\]\[x^{\frac{3}{2}}=\frac{\frac{43}{2}-q}{p}\]\[x=\left(\frac{\frac{43}{2}-q}{p}\right)^{\frac{2}{3}}\]So just put p=2 and q=3 into that equation to solve for one of the values of x.

Answer 4

so i have x^3/2 = 9.25

Answer 5

Yep.

Answer 6

cool :) thanks :D