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.