What are the solutions to the equation (7+x)^2+(7-x)^2=130 ?

The solutions to the equation (7+x)^2+(7-x)^2=130 are x=4,x=-4

Find the zeros of (7+x)^2+(7-x)^2=130

The zeros of (7+x)^2+(7-x)^2=130 are x=4,x=-4

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(7 + x)^{2} + (7 - x)^{2} = 130

x = 4, x = - 4

Solution steps

(7 + x)^{2} + (7 - x)^{2} = 130

Expand

(7 + x)^{2} + (7 - x)^{2} : 2 x^{2} + 98

2 x^{2} + 98 = 130

Move

130

to the left side

2 x^{2} - 32 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 0 \pm 0 ^{2} - 4 \cdot 2 ( - 32 )}{2 \cdot 2}

0^{2} - 4 \cdot 2 (- 32) = 16

x_{1, 2} = \frac{- 0 \pm 16}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 0 + 16}{2 \cdot 2}, x_{2} = \frac{- 0 - 16}{2 \cdot 2}

x = \frac{- 0 + 16}{2 \cdot 2} : 4

x = \frac{- 0 - 16}{2 \cdot 2} : - 4

The solutions to the quadratic equation are:

x = 4, x = - 4

\frac{6}{4 + 5}

\frac{3 - 6}{3 + 6}

x^{2} = - 3

\frac{5 - x}{x - 6} \geq 0

- 1 r < - 2

What are the solutions to the equation (7+x)^2+(7-x)^2=130 ?
The solutions to the equation (7+x)^2+(7-x)^2=130 are x=4,x=-4
Find the zeros of (7+x)^2+(7-x)^2=130
The zeros of (7+x)^2+(7-x)^2=130 are x=4,x=-4