What are the solutions to the equation (x+2)^2+x^2=10^2 ?

The solutions to the equation (x+2)^2+x^2=10^2 are x=6,x=-8

Find the zeros of (x+2)^2+x^2=10^2

The zeros of (x+2)^2+x^2=10^2 are x=6,x=-8

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(x + 2)^{2} + x^{2} = 1 0^{2}

x = 6, x = - 8

Solution steps

(x + 2)^{2} + x^{2} = 1 0^{2}

Expand

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

Expand

1 0^{2} : 100

2 x^{2} + 4 x + 4 = 100

Move

100

to the left side

2 x^{2} + 4 x - 96 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 \cdot 2 ( - 96 )}{2 \cdot 2}

4^{2} - 4 \cdot 2 (- 96) = 28

x_{1, 2} = \frac{- 4 \pm 28}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 4 + 28}{2 \cdot 2}, x_{2} = \frac{- 4 - 28}{2 \cdot 2}

x = \frac{- 4 + 28}{2 \cdot 2} : 6

x = \frac{- 4 - 28}{2 \cdot 2} : - 8

The solutions to the quadratic equation are:

x = 6, x = - 8

- 3 x - 32 \leq 4 x + 24

- \frac{π}{40} + 2 π

∣ x - 8 ∣ \geq 6

r^{2} - 3 r - 28

27 p^{3} - 8

What are the solutions to the equation (x+2)^2+x^2=10^2 ?
The solutions to the equation (x+2)^2+x^2=10^2 are x=6,x=-8
Find the zeros of (x+2)^2+x^2=10^2
The zeros of (x+2)^2+x^2=10^2 are x=6,x=-8