What are the solutions to the equation (x+1)^2+(-2x+8)^2=40 ?

The solutions to the equation (x+1)^2+(-2x+8)^2=40 are x=5,x=1

Find the zeros of (x+1)^2+(-2x+8)^2=40

The zeros of (x+1)^2+(-2x+8)^2=40 are x=5,x=1

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(x + 1)^{2} + (- 2 x + 8)^{2} = 40

x = 5, x = 1

Solution steps

(x + 1)^{2} + (- 2 x + 8)^{2} = 40

Expand

(x + 1)^{2} + (- 2 x + 8)^{2} : 5 x^{2} - 30 x + 65

5 x^{2} - 30 x + 65 = 40

Move

40

to the left side

5 x^{2} - 30 x + 25 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 30 ) \pm ( - 30 ) ^{2} - 4 \cdot 5 \cdot 25}{2 \cdot 5}

(- 30)^{2} - 4 \cdot 5 \cdot 25 = 20

x_{1, 2} = \frac{- ( - 30 ) \pm 20}{2 \cdot 5}

Separate the solutions

x_{1} = \frac{- ( - 30 ) + 20}{2 \cdot 5}, x_{2} = \frac{- ( - 30 ) - 20}{2 \cdot 5}

x = \frac{- ( - 30 ) + 20}{2 \cdot 5} : 5

x = \frac{- ( - 30 ) - 20}{2 \cdot 5} : 1

The solutions to the quadratic equation are:

x = 5, x = 1

625 m^{5} - 256 m n^{4}

2 4^{2} + 7^{2} = c^{2}

32 m^{2} + 28 m - 49

(1 - 82) (1 + 82)

\frac{16}{25}

What are the solutions to the equation (x+1)^2+(-2x+8)^2=40 ?
The solutions to the equation (x+1)^2+(-2x+8)^2=40 are x=5,x=1
Find the zeros of (x+1)^2+(-2x+8)^2=40
The zeros of (x+1)^2+(-2x+8)^2=40 are x=5,x=1