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

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

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

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

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

x = 5, x = 0

Solution steps

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

Expand

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

Expand

(2 x + 2)^{2} + x^{2} : 5 x^{2} + 8 x + 4

9 x^{2} - 12 x + 4 = 5 x^{2} + 8 x + 4

Move

4

to the left side

9 x^{2} - 12 x = 5 x^{2} + 8 x

Move

8 x

to the left side

9 x^{2} - 20 x = 5 x^{2}

Move

5 x^{2}

to the left side

4 x^{2} - 20 x = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 20 ) \pm ( - 20 ) ^{2} - 4 \cdot 4 \cdot 0}{2 \cdot 4}

(- 20)^{2} - 4 \cdot 4 \cdot 0 = 20

x_{1, 2} = \frac{- ( - 20 ) \pm 20}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- ( - 20 ) + 20}{2 \cdot 4}, x_{2} = \frac{- ( - 20 ) - 20}{2 \cdot 4}

x = \frac{- ( - 20 ) + 20}{2 \cdot 4} : 5

x = \frac{- ( - 20 ) - 20}{2 \cdot 4} : 0

The solutions to the quadratic equation are:

x = 5, x = 0

2 m^{2} - 8 m = 0

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

x, y = \frac{x ^{2}}{4}

(x - 4)^{2} + 3 (x + 1) (x - 2) = 8

x, f (x + 2) = x^{2} - x

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