Math Help - Quadratic Equation

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!

2. Originally Posted by BabyMilo
3*y^(1/2)+8*y^(-1/2)-10=0
Multiply through by sqrt[y] to get:

. . . . . $3y\, +\, 8\, -\, 10\sqrt{y}\, =\, 0$

This is a quadratic in $\sqrt{y}$:

. . . . . $3\left(\sqrt{y}\right)^2\, -\, 10\sqrt{y}\, +\, 8\, =\, 0$

Factor this as you would factor 3x^2 - 10x + 8:

. . . . . $(3\sqrt{y}\, -\, 4)(\sqrt{y}\, -\, 2)\, =\, 0$

Then:

. . . . . $\sqrt{y}\, =\, \frac{4}{3}\, \mbox{ or }\, \sqrt{y}\, =\, 2$

Can you take it from there?

3. 3*y^(1/2)+8*y^(-1/2)-10=0
put y^(1/2) = t

3t + 8/t -10 = 0
3t^2 -10t +8 = 0
(3t - 4)(t-2) = 0

t = 4/3 or 2
t^2 = 16/9 or 4
y = 16/9 or 4

4. thanks both of you but does anyone know why my method has gone wrong?
thanks

5. You set sqrt[y] = t, but substituted t^2 for sqrt[y].

6. lol i see where i went wrong now!
thanks!