Solve y
3*y^(1/2)+8*y^(-1/2)-10=0
the answer is 16/9 and 4.
what i did:
root y = t
y= t^2
3t^2+8/t^2-10=0
3t^4-10t^2+8=0
(3t^2-4)(t^2-2)=0
t=16/9 or root 2
which is not what i want.
i want t^2=16/9 and 4
thanks for helping!
Multiply through by sqrt[y] to get:
. . . . .$\displaystyle 3y\, +\, 8\, -\, 10\sqrt{y}\, =\, 0$
This is a quadratic in $\displaystyle \sqrt{y}$:
. . . . .$\displaystyle 3\left(\sqrt{y}\right)^2\, -\, 10\sqrt{y}\, +\, 8\, =\, 0$
Factor this as you would factor 3x^2 - 10x + 8:
. . . . .$\displaystyle (3\sqrt{y}\, -\, 4)(\sqrt{y}\, -\, 2)\, =\, 0$
Then:
. . . . .$\displaystyle \sqrt{y}\, =\, \frac{4}{3}\, \mbox{ or }\, \sqrt{y}\, =\, 2$
Can you take it from there?