Solve for x From the Inequation
Algebra Math Problem: Solve for x when cube root of 3x – 2 is less than x
When x is a real number, Solve for x from the inequation “cube root of 3x – 2 is less than x”.
\sqrt[3]{3x-2}< x
Solving the inequation
\sqrt[3]{3x-2}< x
Take cube in both sides then
3x − 2 < x³
Rearrange this equation then
x³ − 3x + 2 > 0
Factorise the cubic inequation x³ – 3x + 3 > 0, then
x³ – 3x + 2 = x³ – x – 2x + 2
⇒ x³ – 3x + 2 = x(x² – 1) – 2(x – 1)
We know that x² – 1 = (x – 1)(x+1)
⇒ x³ – 3x + 2 = x(x – 1)(x+1) – 2(x – 1)
⇒ x³ – 3x + 2 = (x – 1)(x(x+1) – 2)
So, x³ – 3x + 2 = (x – 1)(x² + x – 2)
Factorise x² + x – 2
⇒ x³ – 3x + 2 = (x – 1)(x² – x + 2x – 2)
⇒ x³ – 3x + 2 = (x – 1)(x(x – 1) + 2(x – 1))
So, x³ – 3x + 2 = (x – 1)(x – 1)(x + 2)
⇒ x³ – 3x + 2 = (x – 1)² (x + 2)
So we can write
(x – 1)² (x + 2) > 0
Where (x – 1)² is greater than 0 (because squares are always positive for real numbers)
Then, (x + 2) > 0
⇒ x > -2
So the solution to the math problem is x > -2