# Irrational equation explanation

• Feb 9th 2006, 03:41 AM
DenMac21
Irrational equation explanation
I have a problem again with irrational equation.

I have to solve this equation:
$\sqrt[3]{x} + \sqrt[3]{{2x - 3}} = \sqrt[3]{{12(x - 1)}}$

I have solve it like this:
$\left( {\sqrt[3]{x} + \sqrt[3]{{2x - 3}}} \right)^3 = \left( {\sqrt[3]{{12(x - 1)}}} \right)^3$
$x + 3\sqrt[3]{x}\sqrt[3]{{2x - 3}}(\sqrt[3]{x} + \sqrt[3]{{2x - 3}}) + 2x - 3 = 12x - 12$
$\sqrt[3]{x}\sqrt[3]{{2x - 3}}(\sqrt[3]{x} + \sqrt[3]{{2x - 3}}) = 3x - 3$
$\sqrt[3]{x}\sqrt[3]{{2x - 3}}\sqrt[3]{{12(x - 1)}} = 3x - 3$
$\sqrt[3]{{x(2x - 3)(12x - 12)}} = 3x - 3$
$x(2x - 3)4(3x - 3) = (3x - 3)^3$
$4x(2x - 3) = (3x - 3)^2$
$8x^2 - 12x = 9x^2 - 18x + 9$
$x^2 - 6x + 9 = 0$
$(x - 3)^2 = 0$
$x = 3$

This is one solution but there is another one that is not obvious: $x=1$.

Now, I have come to that solution by assuming that right side of equation is 0: $\sqrt[3]{{12(x - 1)}}=0$

Can someone explain me how to know exactly what are ALL solutions of irrational equations?
• Feb 9th 2006, 06:08 AM
earboth
Quote:

Originally Posted by DenMac21
I have a problem again with irrational equation.

I have to solve this equation:
$\sqrt[3]{x} + \sqrt[3]{{2x - 3}} = \sqrt[3]{{12(x - 1)}}$

I have solve it like this:
[...snip...]

$x(2x - 3)4(3x - 3) = (3x - 3)^3$
$4x(2x - 3) = (3x - 3)^2$

Hello,

I've quoted those lines where you lost your 2nd solution.
$x(2x - 3)4(3x - 3) - (3x - 3)^3 =0$
$(3x-3) \cdot \left( 4x(2x - 3) - (3x - 3)^2 \right) =0$

A product of two factors is zero, if one factor is zero:
So you get:

$(3x-3)=0 \ \vee \ 4x(2x - 3) - (3x - 3)^2 =0$

After solving both equations you've got all possible results.

Bye
• Feb 9th 2006, 07:15 AM
DenMac21
Quote:

Originally Posted by earboth
Hello,

I've quoted those lines where you lost your 2nd solution.
$x(2x - 3)4(3x - 3) - (3x - 3)^3 =0$
$(3x-3) \cdot \left( 4x(2x - 3) - (3x - 3)^2 \right) =0$

A product of two factors is zero, if one factor is zero:
So you get:

$(3x-3)=0 \ \vee \ 4x(2x - 3) - (3x - 3)^2 =0$

After solving both equations you've got all possible results.

Bye

Thanks :o