Question 208735
To solve equations like this we need to get the "x" out of the square root(s). To eliminate the square roots we need to square each side of the equation.<br>
You could go ahead and square both sides of the original equation but it is easier to if the two square roots are on opposite sides of the equation. So I will start by adding {{{sqrt(x+1)}}} to both sides:
{{{sqrt(2x+3) - sqrt(x+1) + sqrt(x+1) = 1 + sqrt(x+1)}}}
{{{sqrt(2x+3) = 1 + sqrt(x+1)}}}
Now we'll square both sides. (Remember, whenever you square both sides of an equation you may be introducing what are called "extraneous solutions". This means that once we are finished we <b>must</b> check our answers to see if they actually work.)
{{{(sqrt(2x+3))^2 = (1 + sqrt(x+1))^2}}}
Remember that exponents do not distribute. So on the right side we will need to either multiply {{{(1 + sqrt(x+1))(1 + sqrt(x+1))}}} or use the binomial square pattern: {{{(a + b)^2 = a^2 + 2ab + b^2}}}. What we end up with is:
{{{2x + 3 = 1^2 + 2(1)*sqrt(x+1) + (sqrt(x+2))^2}}}
{{{2x + 3 = 1 + 2*sqrt(x+1) + x + 1}}}
{{{2x + 3 = 2*sqrt(x+1) + x + 2}}}
Notice that we eliminated one but not both square roots. To get rid of the remaining square root, isolate it first. So we'll subtract (x+2) from both sides:
{{{x + 1 = 2*sqrt(x+1)}}}
Now we'll square both sides:
{{{(x + 1)^2 = (2(sqrt(x+1)))^2}}}
{{{x^2 + 2(x)(1) + 1^2 = 4(x+1)}}}
{{{x^2 + 2x + 1 = 4x + 4}}}
Our square roots are finally gone. Now we can solve this (quadratic equation). Get one side equal to zero. Subtract (4x+4) from both sides:
{{{x^2 - 2x - 3 = 0}}}
This will factor so we'll solve it that way (instead of using the quadratic equation).
{{{(x - 3)(x + 1) = 0}}}
In order for this product to be zero, one of the factors must be zero.
{{{x - 3) = 0}}} or {{{(x+1) = 0}}}
Now we have two simple equaitons to solve. Add 3 to both sides of the first equation and subtract 1 from both sides of the second equation:
{{{x = 3}}} or {{{x = -1}}}
These are the only possible answers. But since we squared both sides (which we did twice!) we need to make sure they actually work.
Checking x = 3 in the original equation:
{{{sqrt(2x+3) - sqrt(x+1) = 1}}}
{{{sqrt(2(3)+3) - sqrt((3)+1) = 1}}}
{{{sqrt(6+3) - sqrt(4) = 1}}}
{{{sqrt(9) - sqrt(4) = 1}}}
{{{3 - 2 = 1}}}
So x=3 works. It is a solution.
Checking x = -1 in the original equation:
{{{sqrt(2x+3) - sqrt(x+1) = 1}}}
{{{sqrt(2(-1)+3) - sqrt((-1)+1) = 1}}}
{{{sqrt(-2+3) - sqrt(0) = 1}}}
{{{sqrt(1) - sqrt(0) = 1}}}
{{{1 - 0 = 1}}}
x = -1 works, too. It is also a solution.