Answer
$x = 7$
Work Step by Step
We don't want to work with radicals, so we need to transform them. To get rid of radicals, we square both sides of the equation:
$(\sqrt {x - 3})^2 = (x - 5)^2$
Squaring the radical will leave just the binomial:
$x - 3 = (x - 5)^2$
We turn our attention to the right side of the equation to expand the binomial:
$x - 3 = (x - 5)(x - 5)$
Let's use the FOIL method to expand the right side of the equation. With the FOIL method, we multiply the first terms first, then the outer terms, then the inner terms, and, finally, the last terms:
$x - 3 = (x)(x) - 5x - 5x + (-5)(-5)$
Multiply to simplify:
$x - 3 = x^2 - 10x + 25$
Have all terms on one side and zero on the other side:
$0=x^2 - 10x + 25 - x + 3\\
0=x^2-11x+28\\
x^2-11x+28=0$
We solve by factoring.
To factor a polynomial in the form $ax^2 + bx + c$, we look at factors of $c$ such that, when added together equals $b$.
For the trinomial $x^2 - 11x + 28 = 0$, $c=28$so we look for factors of $28$ that when added together will equal $b$ or $-11$. Both factors must be negative. This is because a negative number multiplied with a negative number will equal a positive number; however, when a negative number is added to a negative number, the result will be a negative number.
We came up with the following possibilities:
$(a)(c)$ = $(-28)(-1)$
$b = -29$
$(a)(c)$ = $(-7)(-4)$
$b = -11$
$(a)(c)$ = $(-14)(-2)$
$b = -16$
The second pair works so the factored form of the trinomial is $(x - 7)(x - 4)$.
Thus, the equation above is equivalent to:
$$(x-7)(x-4)=0$$
The Zero-Product property states that if the product of two factors equals zero, then either one of the factors is zero or both factors equal zero. We can, therefore, set each factor to $0$, then solve each equation.
First factor:
$x - 7 = 0$
$x = 7$
Second factor:
$x - 4 = 0$
$x = 4$
To check if our solutions are correct, we plug our solutions back into the original equation to see if the left and right sides equal one another.
Let's plug in $x = 7$ first:
$\sqrt {7 - 3} = 7 - 5$
Evaluate what is in brackets first, according to order of operations:
$\sqrt {4} = 7 - 5$
Evaluate square root:
$2 = 7 - 5$
Now subtract:
$2 = 2$
The left and right sides are equal; therefore, this solution is correct.
Let's check $x = 4$:
$\sqrt {4 - 3} = 4 - 5$
Evaluate what is in brackets first, according to order of operations:
$\sqrt {1} = 4 - 5$
Evaluate square root:
$1 = 4 - 5$
Now subtract:
$1 \ne -1$
The left and right sides are not equal; therefore, this is an extraneous solution.
The solution is $x = 7$.
