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

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

Find the zeros of factor (2x-3)^2=(x+1)(2x-3)

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

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

x = \frac{3}{2}, x = 4

Decimal Notation

x = 1.5, x = 4

Solution steps

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

Expand

(2 x - 3)^{2} : 4 x^{2} - 12 x + 9

Expand

(x + 1) (2 x - 3) : 2 x^{2} - x - 3

4 x^{2} - 12 x + 9 = 2 x^{2} - x - 3

Move

3

to the left side

4 x^{2} - 12 x + 12 = 2 x^{2} - x

Move

x

to the left side

4 x^{2} - 11 x + 12 = 2 x^{2}

Move

2 x^{2}

to the left side

2 x^{2} - 11 x + 12 = 0

Factor

2 x^{2} - 11 x + 12 : (2 x - 3) (x - 4)

(2 x - 3) (x - 4) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

2 x - 3 = 0 or x - 4 = 0

Solve

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

Solve

x - 4 = 0 : x = 4

The solutions to the quadratic equation are:

x = \frac{3}{2}, x = 4

4 x^{2} - 14 x + 2 = 0

0.3 q^{2} + q - 40 = 0

4 d^{2} + 9 d - 7 = 22

- x^{2} + 10 x

x, y^{2} - x y + x^{2} = 8

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