Answer
$x = 1$
Work Step by Step
First, we want to isolate the more complex radical:
$\sqrt {5 - x} = \sqrt {x} + 1$
Square both sides of the equation to eliminate the radical on the left side:
$5 - x = (\sqrt {x} + 1)^2$
Use the the formula $(a+b)^2=a^2+2ab+b^2$ to expand the right side:
$5 - x = x + 2\sqrt {x} + 1$
Move all terms except the radical to the left side of the equation:
$5-x-x-1=2\sqrt{x}\\
4 - 2x = 2\sqrt {x}$
Square both sides of the equation to get rid of the radical:
$(4 - 2x)^2 = (2\sqrt {x})^2$
$(4 - 2x)(4-2x) = 2^2\cdot(\sqrt {x})^2\\
(4-2x)(4-2x)=4x$
Use the FOIL method to distribute on the left side of the equation:
$16 - 8x + 8x + 4x^2 = 4x$
$16 - 16x + 4x^2 = 4x\\
16 - 16x + 4x^2-4x =0\\
16-20x+4x^2=0\\
4x^2-20x+16=0$
Divide both sides of the equation by $4$ to simplify:
$x^2 - 5x + 4 = 0$
We have a quadratic equation, which is in the form $ax^2 + bx + c = 0$. We need to find which factors multiplied together will equal $ac$ but when added together will equal $b$.
In this equation, $ac$ is $4$ and $b$ is $-5$. The factors $-4$ and $-1$ will work.
Let's rewrite the equation in factored form:
$(x - 4)(x - 1) = 0$
Use the Zero-Product Property by equating each factor to $0$< then solve each equation:
First factor:
$x - 4= 0$
$x = 4$
Second factor:
$x - 1 = 0$
$x = 1$
To check if we have any extraneous solutions, we substitute each solution into the original equation to see if the two sides equal one another.
Let's check the first solution:
$\sqrt {5 - 4} - \sqrt {4} = 1$
Combine the terms inside the radicand:
$\sqrt {1} - \sqrt {4} = 1$
Take the square roots:
$1 - 2 = 1$
Combine like terms:
$-1 = 1$
We can see that the two sides are not equal; therefore, this is an extraneous solution.
Let's try the solution $x = 1$:
$\sqrt {5 - 1} - \sqrt {1} = 1$
Combine the terms inside the radicand:
$\sqrt {4} - \sqrt {1} = 1$
Take the square roots:
$2 - 1 = 1$
Combine like terms:
$1 = 1$
Both sides are equal; therefore, this solution is valid.
