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

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

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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

Decimal Notation

x = 4, x = 1.5

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

9

to the right side

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

Move

x

to the left side

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

Move

2 x^{2}

to the left side

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

Divide both sides by

2

\frac{2 x ^{2} - 11 x}{2} = \frac{- 12}{2}

x^{2} - \frac{11 x}{2} = - 6

Rewrite in the form

(x + a)^{2} = b

(x - \frac{11}{4})^{2} = \frac{25}{16}

For

the solutions are

f (x) = a, - a

Solve

x - \frac{11}{4} = \frac{25}{16} : x = 4

Solve

x - \frac{11}{4} = - \frac{25}{16} : x = \frac{3}{2}

The solutions to the quadratic equation are:

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

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

3 y^{2} + 6 = 11 y

y, 2 x^{3} - 7 x^{2} y + 2 y^{2} = q

x^{2} + 11 x + 16 = 0

x, 5 x^{2} - 8 x y + 5 y^{2} - 12 = 0

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