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

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

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

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

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

x = 4, x = 0

Solution steps

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

Expand

3^{2} + x^{2} : 9 + x^{2}

Expand

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

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

Switch sides

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

Move

x^{2}

to the left side

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

Move

9

to the left side

3 x^{2} - 12 x = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 12 ) \pm ( - 12 ) ^{2} - 4 \cdot 3 \cdot 0}{2 \cdot 3}

(- 12)^{2} - 4 \cdot 3 \cdot 0 = 12

x_{1, 2} = \frac{- ( - 12 ) \pm 12}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- ( - 12 ) + 12}{2 \cdot 3}, x_{2} = \frac{- ( - 12 ) - 12}{2 \cdot 3}

x = \frac{- ( - 12 ) + 12}{2 \cdot 3} : 4

x = \frac{- ( - 12 ) - 12}{2 \cdot 3} : 0

The solutions to the quadratic equation are:

x = 4, x = 0

2 + 2 x = - 2

3 x^{2} - 39 = 0

\frac{2 x}{2} + 4 - 4 x = 6 + 2 x

\frac{( x + 13 ) ( x - 15 )}{x + 11} \leq 0

3 x^{3} + 12 x^{2} + 9 x

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