What are the solutions to the equation (1-3x)^4-5(1-3x)^2+4=0 ?

The solutions to the equation (1-3x)^4-5(1-3x)^2+4=0 are x=0,x=1,x=-1/3 ,x= 2/3

Find the zeros of (1-3x)^4-5(1-3x)^2+4=0

The zeros of (1-3x)^4-5(1-3x)^2+4=0 are x=0,x=1,x=-1/3 ,x= 2/3

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(1 - 3 x)^{4} - 5 (1 - 3 x)^{2} + 4 = 0

x = 0, x = 1, x = - \frac{1}{3}, x = \frac{2}{3}

Decimal Notation

x = 0, x = 1, x = - 0.33333 \dots, x = 0.66666 \dots

Solution steps

(1 - 3 x)^{4} - 5 (1 - 3 x)^{2} + 4 = 0

Expand

(1 - 3 x)^{4} - 5 (1 - 3 x)^{2} + 4 : 81 x^{4} - 108 x^{3} + 9 x^{2} + 18 x

81 x^{4} - 108 x^{3} + 9 x^{2} + 18 x = 0

Factor

81 x^{4} - 108 x^{3} + 9 x^{2} + 18 x : 9 x (x - 1) (3 x + 1) (3 x - 2)

9 x (x - 1) (3 x + 1) (3 x - 2) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x = 0 or x - 1 = 0 or 3 x + 1 = 0 or 3 x - 2 = 0

Solve

x - 1 = 0 : x = 1

Solve

3 x + 1 = 0 : x = - \frac{1}{3}

Solve

3 x - 2 = 0 : x = \frac{2}{3}

The solutions are

x = 0, x = 1, x = - \frac{1}{3}, x = \frac{2}{3}

0 = 4 x^{3} + 16 x - 20

x^{3} - 6 x^{2} + 3 x + 1 = 0

y^{3} + 4 y^{2} - 36 y = 144

x^{3} - 3 x^{2} - 2 x - 6 = 0

x^{3} - 8 x^{2} + 20 x - 16 = 0

What are the solutions to the equation (1-3x)^4-5(1-3x)^2+4=0 ?
The solutions to the equation (1-3x)^4-5(1-3x)^2+4=0 are x=0,x=1,x=-1/3 ,x= 2/3
Find the zeros of (1-3x)^4-5(1-3x)^2+4=0
The zeros of (1-3x)^4-5(1-3x)^2+4=0 are x=0,x=1,x=-1/3 ,x= 2/3