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

The solutions to the equation x^2+1=4x are x=2+sqrt(3),x=2-sqrt(3)

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

The zeros of x^2+1=4x are x=2+sqrt(3),x=2-sqrt(3)

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x^{2} + 1 = 4 x

x = 2 + 3, x = 2 - 3

Solution steps

x^{2} + 1 = 4 x

Move

4 x

to the left side

x^{2} + 1 - 4 x = 0

Write in the standard form

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

x^{2} - 4 x + 1 = 0

Solve with the quadratic formula

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

(- 4)^{2} - 4 \cdot 1 \cdot 1 = 23

x_{1, 2} = \frac{- ( - 4 ) \pm 2 3}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 4 ) + 2 3}{2 \cdot 1}, x_{2} = \frac{- ( - 4 ) - 2 3}{2 \cdot 1}

x = \frac{- ( - 4 ) + 2 3}{2 \cdot 1} : 2 + 3

x = \frac{- ( - 4 ) - 2 3}{2 \cdot 1} : 2 - 3

The solutions to the quadratic equation are:

x = 2 + 3, x = 2 - 3

19 - x = 7 x + 11

8 a^{3} - 6 a^{2}

w^{16}

4 (1 + 0.5 m) = 7 m

(f \cdot g) (x)

What are the solutions to the equation x^2+1=4x ?
The solutions to the equation x^2+1=4x are x=2+sqrt(3),x=2-sqrt(3)
Find the zeros of x^2+1=4x
The zeros of x^2+1=4x are x=2+sqrt(3),x=2-sqrt(3)