What are the solutions to the equation 8x^2=(x-5)^2 ?

The solutions to the equation 8x^2=(x-5)^2 are x=-(5(1+2sqrt(2)))/7 ,x=(5(2sqrt(2)-1))/7

Find the zeros of 8x^2=(x-5)^2

The zeros of 8x^2=(x-5)^2 are x=-(5(1+2sqrt(2)))/7 ,x=(5(2sqrt(2)-1))/7

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

x = - \frac{5 ( 1 + 2 2 )}{7}, x = \frac{5 ( 2 2 - 1 )}{7}

Solution steps

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

Expand

(x - 5)^{2} : x^{2} - 10 x + 25

8 x^{2} = x^{2} - 10 x + 25

Switch sides

x^{2} - 10 x + 25 = 8 x^{2}

Move

8 x^{2}

to the left side

- 7 x^{2} - 10 x + 25 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 10 ) \pm ( - 10 ) ^{2} - 4 ( - 7 ) \cdot 25}{2 ( - 7 )}

(- 10)^{2} - 4 (- 7) \cdot 25 = 202

x_{1, 2} = \frac{- ( - 10 ) \pm 20 2}{2 ( - 7 )}

Separate the solutions

x_{1} = \frac{- ( - 10 ) + 20 2}{2 ( - 7 )}, x_{2} = \frac{- ( - 10 ) - 20 2}{2 ( - 7 )}

x = \frac{- ( - 10 ) + 20 2}{2 ( - 7 )} : - \frac{5 ( 1 + 2 2 )}{7}

x = \frac{- ( - 10 ) - 20 2}{2 ( - 7 )} : \frac{5 ( 2 2 - 1 )}{7}

The solutions to the quadratic equation are:

x = - \frac{5 ( 1 + 2 2 )}{7}, x = \frac{5 ( 2 2 - 1 )}{7}

(2 - (4))^{2} + (1 - (- 3))^{2} = r^{2}

2 x^{2} + 17 x + 24 = 0

x^{2} - \frac{1}{4} x + 2

4 (6 - 4 x)^{2} = 49

y^{2} + 3 y - 28 = 0

What are the solutions to the equation 8x^2=(x-5)^2 ?
The solutions to the equation 8x^2=(x-5)^2 are x=-(5(1+2sqrt(2)))/7 ,x=(5(2sqrt(2)-1))/7
Find the zeros of 8x^2=(x-5)^2
The zeros of 8x^2=(x-5)^2 are x=-(5(1+2sqrt(2)))/7 ,x=(5(2sqrt(2)-1))/7