What are the solutions to the equation x^2+1=sqrt(6)x ?

The solutions to the equation x^2+1=sqrt(6)x are x=(sqrt(6)+sqrt(2))/2 ,x=(sqrt(6)-sqrt(2))/2

Find the zeros of x^2+1=sqrt(6)x

The zeros of x^2+1=sqrt(6)x are x=(sqrt(6)+sqrt(2))/2 ,x=(sqrt(6)-sqrt(2))/2

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x^{2} + 1 = 6 x

x = \frac{6 + 2}{2}, x = \frac{6 - 2}{2}

Solution steps

x^{2} + 1 = 6 x

Move

6 x

to the left side

x^{2} + 1 - 6 x = 0

Write in the standard form

a x^{2} + b x + c = 0

x^{2} - 6 x + 1 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 6 ) \pm ( - 6 ) ^{2} - 4 \cdot 1 \cdot 1}{2 \cdot 1}

(- 6)^{2} - 4 \cdot 1 \cdot 1 = 2

x_{1, 2} = \frac{- ( - 6 ) \pm 2}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 6 ) + 2}{2 \cdot 1}, x_{2} = \frac{- ( - 6 ) - 2}{2 \cdot 1}

x = \frac{- ( - 6 ) + 2}{2 \cdot 1} : \frac{6 + 2}{2}

x = \frac{- ( - 6 ) - 2}{2 \cdot 1} : \frac{6 - 2}{2}

The solutions to the quadratic equation are:

x = \frac{6 + 2}{2}, x = \frac{6 - 2}{2}

C = 1600 - 10 x + 0.25 x^{2}

(\frac{3 x + 2}{4})^{2} = 12

x + 3 = 4 x^{2}

p^{2} + 6 p - 72 = 0

(x - 6) (2 x + 4) = 0

What are the solutions to the equation x^2+1=sqrt(6)x ?
The solutions to the equation x^2+1=sqrt(6)x are x=(sqrt(6)+sqrt(2))/2 ,x=(sqrt(6)-sqrt(2))/2
Find the zeros of x^2+1=sqrt(6)x
The zeros of x^2+1=sqrt(6)x are x=(sqrt(6)+sqrt(2))/2 ,x=(sqrt(6)-sqrt(2))/2