What are the solutions to the equation x^2-2x+1=16(x-4) ?

The solutions to the equation x^2-2x+1=16(x-4) are x=13,x=5

Find the zeros of x^2-2x+1=16(x-4)

The zeros of x^2-2x+1=16(x-4) are x=13,x=5

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x^{2} - 2 x + 1 = 16 (x - 4)

x = 13, x = 5

Solution steps

x^{2} - 2 x + 1 = 16 (x - 4)

Expand

16 (x - 4) : 16 x - 64

x^{2} - 2 x + 1 = 16 x - 64

Move

64

to the left side

x^{2} - 2 x + 65 = 16 x

Move

16 x

to the left side

x^{2} - 18 x + 65 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 18 ) \pm ( - 18 ) ^{2} - 4 \cdot 1 \cdot 65}{2 \cdot 1}

(- 18)^{2} - 4 \cdot 1 \cdot 65 = 8

x_{1, 2} = \frac{- ( - 18 ) \pm 8}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 18 ) + 8}{2 \cdot 1}, x_{2} = \frac{- ( - 18 ) - 8}{2 \cdot 1}

x = \frac{- ( - 18 ) + 8}{2 \cdot 1} : 13

x = \frac{- ( - 18 ) - 8}{2 \cdot 1} : 5

The solutions to the quadratic equation are:

x = 13, x = 5

x y^{3} - 9 x y z^{2}

3 (x - 3) = 6

x^{3} - 3 x^{2} - 2 x + 6

(x^{3} y^{3})^{3} \cdot x y^{2}

20 m^{2} + 44 m - 15

What are the solutions to the equation x^2-2x+1=16(x-4) ?
The solutions to the equation x^2-2x+1=16(x-4) are x=13,x=5
Find the zeros of x^2-2x+1=16(x-4)
The zeros of x^2-2x+1=16(x-4) are x=13,x=5