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!
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Originally Posted by BabyMilo 3*y^(1/2)+8*y^(-1/2)-10=0 Multiply through by sqrt[y] to get: . . . . . This is a quadratic in : . . . . . Factor this as you would factor 3x^2 - 10x + 8: . . . . . Then: . . . . . Can you take it from there?
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
thanks both of you but does anyone know why my method has gone wrong? thanks
You set sqrt[y] = t, but substituted t^2 for sqrt[y].
lol i see where i went wrong now! thanks!
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