What are the solutions to the equation (x+3)^2+(x-1)^2=40 ?

The solutions to the equation (x+3)^2+(x-1)^2=40 are x=3,x=-5

Find the zeros of (x+3)^2+(x-1)^2=40

The zeros of (x+3)^2+(x-1)^2=40 are x=3,x=-5

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(x + 3)^{2} + (x - 1)^{2} = 40

x = 3, x = - 5

Solution steps

(x + 3)^{2} + (x - 1)^{2} = 40

Expand

(x + 3)^{2} + (x - 1)^{2} : 2 x^{2} + 4 x + 10

2 x^{2} + 4 x + 10 = 40

Move

40

to the left side

2 x^{2} + 4 x - 30 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 \cdot 2 ( - 30 )}{2 \cdot 2}

4^{2} - 4 \cdot 2 (- 30) = 16

x_{1, 2} = \frac{- 4 \pm 16}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 4 + 16}{2 \cdot 2}, x_{2} = \frac{- 4 - 16}{2 \cdot 2}

x = \frac{- 4 + 16}{2 \cdot 2} : 3

x = \frac{- 4 - 16}{2 \cdot 2} : - 5

The solutions to the quadratic equation are:

x = 3, x = - 5

x^{2} - 8 x - 15 = 0

\frac{8 !}{( 8 - 3 ) ! 3 !}

\frac{2 x - 4}{5} + 6 = 8

- 5.3 \geq 3.6 x

\frac{( 2 + 0.01 ) ^{3} - 8}{0.01}

What are the solutions to the equation (x+3)^2+(x-1)^2=40 ?
The solutions to the equation (x+3)^2+(x-1)^2=40 are x=3,x=-5
Find the zeros of (x+3)^2+(x-1)^2=40
The zeros of (x+3)^2+(x-1)^2=40 are x=3,x=-5