What are the solutions to the equation (x-4)^2-10=(x-1)^2-x^2 ?

The solutions to the equation (x-4)^2-10=(x-1)^2-x^2 are x=5,x=1

Find the zeros of (x-4)^2-10=(x-1)^2-x^2

The zeros of (x-4)^2-10=(x-1)^2-x^2 are x=5,x=1

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(x - 4)^{2} - 10 = (x - 1)^{2} - x^{2}

x = 5, x = 1

Solution steps

(x - 4)^{2} - 10 = (x - 1)^{2} - x^{2}

Expand

(x - 4)^{2} - 10 : x^{2} - 8 x + 6

Expand

(x - 1)^{2} - x^{2} : - 2 x + 1

x^{2} - 8 x + 6 = - 2 x + 1

Move

1

to the left side

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

Move

2 x

to the left side

x^{2} - 6 x + 5 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 6 ) \pm ( - 6 ) ^{2} - 4 \cdot 1 \cdot 5}{2 \cdot 1}

(- 6)^{2} - 4 \cdot 1 \cdot 5 = 4

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

Separate the solutions

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

x = \frac{- ( - 6 ) + 4}{2 \cdot 1} : 5

x = \frac{- ( - 6 ) - 4}{2 \cdot 1} : 1

The solutions to the quadratic equation are:

x = 5, x = 1

x^{2} + x + 3

x^{2} + x + 6

4 w^{2} - 19 w + 21 = 2

y, x^{2} + y^{2} - 4 x + 6 y + 13 = 0

2^{2} + 21^{2} = x^{2}

What are the solutions to the equation (x-4)^2-10=(x-1)^2-x^2 ?
The solutions to the equation (x-4)^2-10=(x-1)^2-x^2 are x=5,x=1
Find the zeros of (x-4)^2-10=(x-1)^2-x^2
The zeros of (x-4)^2-10=(x-1)^2-x^2 are x=5,x=1