Radical Equation Math Help

Question

How do you know what power to raise both sides of an equation to if the indices of the radicals are different? For example, given the equation , what index would you use? Why? Explain your reasoning. Make sure to address this specific example, then provide an additional, unique example that also shows how your reasoning is correct.  What one number could you raise both sides of the equation to that would get rid of a square root (a base to the power of 1/2) on one side of an equation, and a cube root (a base to a power of 1/3) on the other side, all at once? Please Explain why on what index would you use and your reasoning.

Answer 1

Take LCM of denominators of radical

For e.g. 2,3 has LCM = 6

So raise both sides to power of 6 and then solve.

(x+1)^1/2*6 = (x-1)^1/3*6

(x+1)³ = (x-1)²

^{Hope this helps}

Answer 2

raise both sides to the 6th power

that eliminates all roots That's the same as squaring the left side and then cubing it,

while cubing the right side, then squaring it

(x+1)^3 = (x-1)^2 or

x^3 + 3x^2 + 3x + 1 = x^2 -2x +1

one's cancel, so there's no constant term

x^3 + 3x^2 + 3x = x^2 -2x

x^3 +2x^2 +5x = 0

factor out an x

x(x^2 + 2x + 5) = 0

x = 0 is one root or zero

use the quadratic equation on the other quadratic factor

x = -2/2 + or - (1/2)sqr(4-4(5))

= -1 + or - 2i

That gives 3 solutions, one real, two imaginary

x = 0, -1 + 2i, and -1 - 2i

once you reached the cubic equation, you know there's 3 possible solutions

all 3 real, or one real & 2 imaginary

cubic equations cross the x-axis at least once, for one real zero

but when you multiply by a variable you may introduce an extraneous solution not in the original problem, so you need to check the answers

square root of 0+1 is 1

but cube root of 0-1 = -1

0 is an extraneous solution not in the original problem

Unless, the square root was intended to include the negative square root, but when it has no sign, it's assumed to include just the positive square root.

It's harder to check the imaginary solutions under cube and square roots

but if there is a solution it's imaginary