square root + cube root of a number

**dugongster** · December 5th 2010, 10:07 AM

The square root and cube root of a number is equal to 36. What is the number ?

Can this really be solved by logarithms ?

I got far as log ( 36 - y raised to 1/2 ) = log y

I don't know what to do next ?

**e^(i*pi)** · December 5th 2010, 10:14 AM

$\sqrt{x} + \sqrt[3]{x} = 36 \: \: , \: x \geq 0$

Logarithms seem like more trouble than they're worth.

Let $u = \sqrt[6]{x} \implies x = u^6$ (6 is the LCD of 2 and 3).

Thus we get $f(u) = u^3 + u^2 - 36 = 0$ . u^3 comes from the rules of surds, namely: $\sqrt{a} = a^{1/2} = a^{3/6} = \sqrt[6]{a^3} = u^3$

Solve the polynomial for u and then it should be trivial to find the value of x but remember to check for extraneous results and that any solution is in the domain

edit 2: Hint f(3) = 0

If you don't mean adding them then please clarify what the question wants

**dugongster** · December 5th 2010, 10:35 AM

You're the man !

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