What are the solutions to the equation (2x^2-7x)^2+2x^2-7x=20 ?

The solutions to the equation (2x^2-7x)^2+2x^2-7x=20 are x=1,x=4,x=-1/2 ,x= 5/2

Find the zeros of (2x^2-7x)^2+2x^2-7x=20

The zeros of (2x^2-7x)^2+2x^2-7x=20 are x=1,x=4,x=-1/2 ,x= 5/2

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(2 x^{2} - 7 x)^{2} + 2 x^{2} - 7 x = 20

x = 1, x = 4, x = - \frac{1}{2}, x = \frac{5}{2}

Decimal Notation

x = 1, x = 4, x = - 0.5, x = 2.5

Solution steps

(2 x^{2} - 7 x)^{2} + 2 x^{2} - 7 x = 20

Expand

(2 x^{2} - 7 x)^{2} + 2 x^{2} - 7 x : 4 x^{4} - 28 x^{3} + 51 x^{2} - 7 x

4 x^{4} - 28 x^{3} + 51 x^{2} - 7 x = 20

Move

20

to the left side

4 x^{4} - 28 x^{3} + 51 x^{2} - 7 x - 20 = 0

Factor

4 x^{4} - 28 x^{3} + 51 x^{2} - 7 x - 20 : (x - 1) (x - 4) (2 x + 1) (2 x - 5)

(x - 1) (x - 4) (2 x + 1) (2 x - 5) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x - 1 = 0 or x - 4 = 0 or 2 x + 1 = 0 or 2 x - 5 = 0

Solve

x - 1 = 0 : x = 1

Solve

x - 4 = 0 : x = 4

Solve

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

Solve

2 x - 5 = 0 : x = \frac{5}{2}

The solutions are

x = 1, x = 4, x = - \frac{1}{2}, x = \frac{5}{2}

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

(x^{2} + x + 1)^{2} = x^{4} + 8 x^{2} + 1

x^{3} + x^{3} = 2

(x - 1)^{3} = 1

6 x^{3} + 13 x^{2} - 4 = 0

What are the solutions to the equation (2x^2-7x)^2+2x^2-7x=20 ?
The solutions to the equation (2x^2-7x)^2+2x^2-7x=20 are x=1,x=4,x=-1/2 ,x= 5/2
Find the zeros of (2x^2-7x)^2+2x^2-7x=20
The zeros of (2x^2-7x)^2+2x^2-7x=20 are x=1,x=4,x=-1/2 ,x= 5/2