What is the answer to x^2-2x+sqrt(3-x)=sqrt(3-x)+8 ?

The answer to x^2-2x+sqrt(3-x)=sqrt(3-x)+8 is x=-2,x=4

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x^2-2x+sqrt(3-x)=sqrt(3-x)+8

x^{2} - 2 x + 3 - x = 3 - x + 8

x = - 2, x = 4

Solution steps

x^{2} - 2 x + 3 - x = 3 - x + 8

Subtract

3 - x + 8

from both sides

x^{2} - 2 x + 3 - x - (3 - x + 8) = 3 - x + 8 - (3 - x + 8)

Simplify

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

Factor

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

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

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x + 2 = 0 or x - 4 = 0

Solve

x + 2 = 0 : x = - 2

Solve

x - 4 = 0 : x = 4

x = - 2, x = 4

Verify Solutions:

x = - 2

True

, x = 4

True

The solutions are

x = - 2, x = 4

3 - 2 v + 17 = 20

3 x - 1 - 4 = 2

x - \frac{1}{2} x = 0

x^{\frac{3}{2}} - 2 x^{\frac{3}{4}} + 1 = 0

What is the answer to x^2-2x+sqrt(3-x)=sqrt(3-x)+8 ?
The answer to x^2-2x+sqrt(3-x)=sqrt(3-x)+8 is x=-2,x=4
Solve for x: x^2-2x+sqrt(3-x)=sqrt(3-x)+8
The solution is x=-2,x=4