What are the solutions to the equation 4(x^2-x)^2-11(x^2-x)+6=0 ?

The solutions to the equation 4(x^2-x)^2-11(x^2-x)+6=0 are x=-1,x=2,x=-1/2 ,x= 3/2

Find the zeros of 4(x^2-x)^2-11(x^2-x)+6=0

The zeros of 4(x^2-x)^2-11(x^2-x)+6=0 are x=-1,x=2,x=-1/2 ,x= 3/2

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4 (x^{2} - x)^{2} - 11 (x^{2} - x) + 6 = 0

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

Decimal Notation

x = - 1, x = 2, x = - 0.5, x = 1.5

Solution steps

4 (x^{2} - x)^{2} - 11 (x^{2} - x) + 6 = 0

Expand

4 (x^{2} - x)^{2} - 11 (x^{2} - x) + 6 : 4 x^{4} - 8 x^{3} - 7 x^{2} + 11 x + 6

4 x^{4} - 8 x^{3} - 7 x^{2} + 11 x + 6 = 0

Factor

4 x^{4} - 8 x^{3} - 7 x^{2} + 11 x + 6 : (x + 1) (x - 2) (2 x + 1) (2 x - 3)

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

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

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

Solve

x + 1 = 0 : x = - 1

Solve

x - 2 = 0 : x = 2

Solve

2 x + 1 = 0 : x = - \frac{1}{2}

Solve

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

The solutions are

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

5 y^{3} + 33 y^{2} = 14 y

x^{4} + x^{2} - 2 x = 0

20 x^{3} - 15 x = 3

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

10 = x^{10}

What are the solutions to the equation 4(x^2-x)^2-11(x^2-x)+6=0 ?
The solutions to the equation 4(x^2-x)^2-11(x^2-x)+6=0 are x=-1,x=2,x=-1/2 ,x= 3/2
Find the zeros of 4(x^2-x)^2-11(x^2-x)+6=0
The zeros of 4(x^2-x)^2-11(x^2-x)+6=0 are x=-1,x=2,x=-1/2 ,x= 3/2