What are the solutions to the equation 5^2=4x^2-4x+1+x^2 ?

The solutions to the equation 5^2=4x^2-4x+1+x^2 are x=(2(1+sqrt(31)))/5 ,x=(2(1-sqrt(31)))/5

Find the zeros of 5^2=4x^2-4x+1+x^2

The zeros of 5^2=4x^2-4x+1+x^2 are x=(2(1+sqrt(31)))/5 ,x=(2(1-sqrt(31)))/5

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5^{2} = 4 x^{2} - 4 x + 1 + x^{2}

x = \frac{2 ( 1 + 31 )}{5}, x = \frac{2 ( 1 - 31 )}{5}

Decimal Notation

x = 2.62710 \dots, x = - 1.82710 \dots

Solution steps

5^{2} = 4 x^{2} - 4 x + 1 + x^{2}

Expand

5^{2} : 25

25 = 5 x^{2} - 4 x + 1

Switch sides

5 x^{2} - 4 x + 1 = 25

Move

25

to the left side

5 x^{2} - 4 x - 24 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 5 ( - 24 )}{2 \cdot 5}

(- 4)^{2} - 4 \cdot 5 (- 24) = 431

x_{1, 2} = \frac{- ( - 4 ) \pm 4 31}{2 \cdot 5}

Separate the solutions

x_{1} = \frac{- ( - 4 ) + 4 31}{2 \cdot 5}, x_{2} = \frac{- ( - 4 ) - 4 31}{2 \cdot 5}

x = \frac{- ( - 4 ) + 4 31}{2 \cdot 5} : \frac{2 ( 1 + 31 )}{5}

x = \frac{- ( - 4 ) - 4 31}{2 \cdot 5} : \frac{2 ( 1 - 31 )}{5}

The solutions to the quadratic equation are:

x = \frac{2 ( 1 + 31 )}{5}, x = \frac{2 ( 1 - 31 )}{5}

3 x^{2} - 5 x + 10 = 0

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

5 x^{2} - 5 x - 2 = 0

x^{2} + 3 x - 4 = 50

3 x^{2} - 12 = x^{2} - 44

What are the solutions to the equation 5^2=4x^2-4x+1+x^2 ?
The solutions to the equation 5^2=4x^2-4x+1+x^2 are x=(2(1+sqrt(31)))/5 ,x=(2(1-sqrt(31)))/5
Find the zeros of 5^2=4x^2-4x+1+x^2
The zeros of 5^2=4x^2-4x+1+x^2 are x=(2(1+sqrt(31)))/5 ,x=(2(1-sqrt(31)))/5