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C. Solve: x + square root of 5-x = 3

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Both x=4 and x=1 are valid solutions to the original equation.

To solve the equation x+ √5−x =3, we can follow these steps:

Move the x term to the other side of the equation:

√5−x =3−x

Square both sides of the equation to eliminate the square root:

5−x=(3−x)^2

Expand and simplify the right side of the equation:

5−x=9−6x+x^2

Move all terms to one side of the equation to set it to zero:

x^2 −5x+4=0

Factor the quadratic expression:

(x−4)(x−1)=0

Set each factor equal to zero and solve for x:

x−4=0⟹x=4

x−1=0⟹x=1

So, the solutions to the equation x+ √5−x =3 are x=4 and x=1.

However, we need to check whether these solutions are valid by ensuring that the expressions inside the square root are non-negative. Let's check:

For x=4:

5−x=5−4=1, which is non-negative.

For x=1:

5−x=5−1=4, which is also non-negative.

Therefore, both x=4 and x=1 are valid solutions to the original equation.

User Gergo
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