Question 624707
Since exponents of 1/2 are square roots, I'm going to do this problem using square roots instead of the fractional exponents.<br>
{{{sqrt(3x+1) - sqrt(x+4) = 1}}}
Here is a procedure for solving square root equations like this:<ol><li>Isolate a square root.</li><li>Square both sides of the equation. Note: The side with the isolated square root should be easy to square. But be careful with the other side.)</li><li>If there is still a square root in the equation, repeat steps 1-3.</li><li>At this point there should no longer be any square roots. Solve this equation using whatever techniques are appropriate.</li><li>Check your solution(s). This is <i>not optional!</i> Whenever you square both sides of an equation (which you have done at least once at step 2) you can end up with what are called extraneous solutions. Extraneous solutions are solutions that fit the squared equation <i>but do not fit the original equation.</i> Extraneous solutions can happen any time you square both sides. They do not mean a mistake was made or that you should never square both sides of an equation. So you must check your solutions in the original equation and reject any solutions which do not work.</li></ol>Let's see this in action:
1. Isolate a square root.
Adding {{{sqrt(x+4)}}} to each side:
{{{sqrt(3x+1) = 1 + sqrt(x+4)}}}<br>
2. Square both sides:
{{{(sqrt(3x+1))^2 = (1 + sqrt(x+4))^2}}}
Squaring the left side is simple. Squaring the right side requires using FOIL on {{{(1 + sqrt(x+4))*(1 + sqrt(x+4))}}} or use the {{{(a+b)^2 = a^2+2ab+b^2}}} pattern with the "a" being "1" and the "b" being {{{sqrt(x+4)}}}. Personally I prefer using patterns:
{{{3x+1 = (1)^2 +2(1)(sqrt(x+4)) + (sqrt(x+4))^2}}}
which simplifies as follows:
{{{3x+1 = 1 +2*sqrt(x+4) + (x+4)}}}
{{{3x+1 = 2*sqrt(x+4) + x+5}}}<br>
3. There is still a square root so we repeat...
1. Isolate a square root.
Subtracting x and 5 from each side:
{{{2x-4 = 2*sqrt(x+4)}}}
The square root is sufficiently isolated. The "2" is not a problem. The right side will still square easily. But if it bothers you then divide both sides by 2.<br>
2. Square both sides.
{{{(2x-4)^2 = (2*sqrt(x+4))^2}}}
{{{(2x)^2 - 2(2x)(4) + (4)^2 = 4(x+4)}}}
which simplifies as follows:
{{{4x^2-16x+16 = 4x+16}}}<br>
3. There are no more square roots so we can proceed.<br>
4. Solve the equation.
Because of the {{{x^2}}} term this is a quadratic equation. So we want one side to be zero. Subtracting 4x and 16 from each side we get:
{{{4x^2-20x=0}}}
Next we factor (or use the Quadratic Formula). This factors very easily:
{{{4x(x-5) = 0}}}
From the Zero Product Property we know that one of the factors must be zero:
4x = 0 or x-5 = 0
Solving these we get:
x = 0 or x = 5<br>
5. Check your solution(s).
Use the original equation:
{{{sqrt(3x+1) - sqrt(x+4) = 1}}}
Checking x = 0:
{{{sqrt(3(0)+1) - sqrt((0)+4) = 1}}}
which simplifies as follows:
{{{sqrt(0+1) - sqrt(0+4) = 1}}}
{{{sqrt(1) - sqrt(4) = 1}}}
{{{1 - 2 = 1}}}
{{{-1 = 1}}} Check fails! This is an extraneous solution which we discard. (Note: This does not mean we made a mistake earlier!)<br>
Checking x = 5:
{{{sqrt(3(5)+1) - sqrt((5)+4) = 1}}}
which simplifies as follows:
{{{sqrt(15+1) - sqrt(5+4) = 1}}}
{{{sqrt(16) - sqrt(9) = 1}}}
{{{4 - 3 = 1}}}
{{{1 = 1}}} Check!
So the only solution to your equation is x = 5.