What are the solutions to the equation (x+7)^2=(2x-1)^2 ?

The solutions to the equation (x+7)^2=(2x-1)^2 are x=-2,x=8

Find the zeros of (x+7)^2=(2x-1)^2

The zeros of (x+7)^2=(2x-1)^2 are x=-2,x=8

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(x + 7)^{2} = (2 x - 1)^{2}

x = - 2, x = 8

Solution steps

(x + 7)^{2} = (2 x - 1)^{2}

Expand

(x + 7)^{2} : x^{2} + 14 x + 49

Expand

(2 x - 1)^{2} : 4 x^{2} - 4 x + 1

x^{2} + 14 x + 49 = 4 x^{2} - 4 x + 1

Move

1

to the left side

x^{2} + 14 x + 48 = 4 x^{2} - 4 x

Move

4 x

to the left side

x^{2} + 18 x + 48 = 4 x^{2}

Move

4 x^{2}

to the left side

- 3 x^{2} + 18 x + 48 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 18 \pm 1 8 ^{2} - 4 ( - 3 ) \cdot 48}{2 ( - 3 )}

1 8^{2} - 4 (- 3) \cdot 48 = 30

x_{1, 2} = \frac{- 18 \pm 30}{2 ( - 3 )}

Separate the solutions

x_{1} = \frac{- 18 + 30}{2 ( - 3 )}, x_{2} = \frac{- 18 - 30}{2 ( - 3 )}

x = \frac{- 18 + 30}{2 ( - 3 )} : - 2

x = \frac{- 18 - 30}{2 ( - 3 )} : 8

The solutions to the quadratic equation are:

x = - 2, x = 8

x^{2} - \frac{5}{3} x

x^{2} = 12 (x - 5)

ab c x^{2} - (a^{2} b^{2} + c^{2}) x + ab c = 0

- 22 - 3 x^{2} - 2 x = - 3

- 3 x^{2} = 12

What are the solutions to the equation (x+7)^2=(2x-1)^2 ?
The solutions to the equation (x+7)^2=(2x-1)^2 are x=-2,x=8
Find the zeros of (x+7)^2=(2x-1)^2
The zeros of (x+7)^2=(2x-1)^2 are x=-2,x=8