Solve the equation 3x^2/3 + x^1/3 − 2 = 0
Let u = x^1/3
The equation can therefore be written as:
3u^2+u-2=0
This can be factorised to:
(3u-2)(u+1)+0
Therefore: u = 2/3 or u = -1 OR x^1/3 = 2/3 or x^1/3 = -1
So: x = (2/3)^3 or x = (-1)^3
x = 8/27 or x = -1
NH
Related Maths A Level answers
All answers ▸The curve C has equation y = 3x^4 – 8x^3 – 3 (a) Find (i) dy/dx (ii) d^2y/dx^2 (3 marks) (b) Verify that C has a stationary point when x = 2 (2marks) (c) Determine the nature of this stationary point, giving a reason for your answer. (2)
Find the Total Area between the curve x^3 -3x^2 +2x and the x-axis, when 0 ≤ x ≤ 2.
Prove that the equation y = 3x^4 - 8x^3 - 3 has a turning point at x=2
a typical question would be a setof parametric equations y(t) and x(t), asking you to find dy/dx and then the tangent/normal to the curve at a certain point (ie t = 2)