What are the solutions to the equation 5x^2=4x^2-8x+1 ?

The solutions to the equation 5x^2=4x^2-8x+1 are x=-4-sqrt(17),x=sqrt(17)-4

Find the zeros of 5x^2=4x^2-8x+1

The zeros of 5x^2=4x^2-8x+1 are x=-4-sqrt(17),x=sqrt(17)-4

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5 x^{2} = 4 x^{2} - 8 x + 1

x = - 4 - 17, x = 17 - 4

Decimal Notation

x = - 8.12310 \dots, x = 0.12310 \dots

Solution steps

5 x^{2} = 4 x^{2} - 8 x + 1

Switch sides

4 x^{2} - 8 x + 1 = 5 x^{2}

Move

5 x^{2}

to the left side

- x^{2} - 8 x + 1 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 8 ) \pm ( - 8 ) ^{2} - 4 ( - 1 ) \cdot 1}{2 ( - 1 )}

(- 8)^{2} - 4 (- 1) \cdot 1 = 217

x_{1, 2} = \frac{- ( - 8 ) \pm 2 17}{2 ( - 1 )}

Separate the solutions

x_{1} = \frac{- ( - 8 ) + 2 17}{2 ( - 1 )}, x_{2} = \frac{- ( - 8 ) - 2 17}{2 ( - 1 )}

x = \frac{- ( - 8 ) + 2 17}{2 ( - 1 )} : - 4 - 17

x = \frac{- ( - 8 ) - 2 17}{2 ( - 1 )} : 17 - 4

The solutions to the quadratic equation are:

x = - 4 - 17, x = 17 - 4

x^{4} + 10 x^{3} + 21 x^{2} - 40 x - 100 > 0

5 x + 3 - 1 = x + 2 + 4 x

n^{3} - 9 n

\frac{4 !}{( 2 - 4 ) ! \cdot 2 !}

5 x - 5 = 3 x + 13

What are the solutions to the equation 5x^2=4x^2-8x+1 ?
The solutions to the equation 5x^2=4x^2-8x+1 are x=-4-sqrt(17),x=sqrt(17)-4
Find the zeros of 5x^2=4x^2-8x+1
The zeros of 5x^2=4x^2-8x+1 are x=-4-sqrt(17),x=sqrt(17)-4