Question 706050
{{{sqrt(2x-1)-1=sqrt(1-x)}}}
Here's a procedure for solving equations like these:<ol><li>Isolate a square root.</li><li>Square both sides of the equation. (Be careful with this step. This is where most mistakes are made. Squaring the isolated square root is as easy as it looks. Mistakes are often made when squaring the other side of the equation.)</li><li>If there are still any square roots remaining, repeat steps 1,2 and 3. (Note: Squaring both sides of an equation correctly does not always eliminate all the square roots. So there may still be square roots after step 2!)</li><li>At this point there should be no square roots left. Use appropriate techniques to solve whatever kind of equation you now have.)</li><li>Check your solution(s). This is <i>not optional</i> when solving these equations. Squaring both sides of an equation, which we have done at least once at step 2, can introduce what are called extraneous solutions. Extraneous solutions are solutions that fit the squared equation but do not fit the original equation. These extraneous solutions can occur <i>even if no mistakes have been made!</i> So you must check to see if your possible solutions actually work.</li></ol>Let's see this in action:
{{{sqrt(2x-1)-1=sqrt(1-x)}}}
1. Isolate a square root.
The square root on the right is already isolated.<br>
2. Square both sides.
{{{(sqrt(2x-1)-1)^2=(sqrt(1-x))^2}}}
The right side squares easily. But we must use FOIL or the {{{(a-b)^2=a^2-2ab+b^2}}} pattern to square the left side correctly. I prefer using the pattern:
{{{(sqrt(2x-1))^2-2*(sqrt(2x-1))(1)+ (1)^2=1-x}}}
Simplifying the left side:
{{{(2x-1)-2sqrt(2x-1)+ 1=1-x}}}
The 1's on the left cancel:
{{{2x-2sqrt(2x-1)=1-x}}}<br>
3. We still have a square root. So back to step 1.
1. Isolate a square root.
There's only one left. Subtracting 2x from each side:
{{{-2sqrt(2x-1)=1-3x}}}
The square root on the left is sufficiently isolated. But if the -2 in front bothers you then divide both sides by -2 to get rid of it.<br>
2. Square both sides.
{{{(-2sqrt(2x-1))^2=(1-3x)^2}}}
The left side is fairly easy to square. We'll use the pattern again on the right side:
{{{4*(2x-1) = (1)^2-2(1)(3x)+(3x)^2}}}
Simplifying:
{{{8x-4 = 1-6x+9x^2}}}<br>
3. The square roots are gone! So on to step 4.<br>
4. Solve the equation.
We now have a quadratic equation. To solve it we want one side to be zero. Subtracting 8x and adding 4 we get:
{{{0 = 9x^2-14x+5}}}
Now we factor:
0 = (9x-5)(x-1)
From the Zero Product Property:
9x-5 = 0 or x-1 = 0
Solving these we get:
x = 5/9 or x = 1<br>
5. Check the solution(s).
Use the original equation to check:
{{{sqrt(2x-1)-1=sqrt(1-x)}}}
Checking x = 5/9:
{{{sqrt(2(5/9)-1)-1=sqrt(1-(5/9))}}}
Simplifying...
{{{sqrt(10/9-1)-1=sqrt(1-5/9)}}}
{{{sqrt(10/9-9/9)-1=sqrt(9/9-5/9)}}}
{{{sqrt(1/9)-1=sqrt(4/9)}}}
{{{1/3-1=2/3}}}
{{{1/3-3/3=2/3}}}
{{{-2/3=2/3}}}
This is not true. This "solution" does not work. It is extraneous and we must reject it as a solution.<br>
Checking x = 1:
{{{sqrt(2(1)-1)-1=sqrt(1-(1))}}}
Simplifying...
{{{sqrt(2-1)-1=sqrt(1-1)}}}
{{{sqrt(1)-1=sqrt(0)}}}
{{{1-1=0}}}
{{{0 = 0}}}
This solution checks out! So the only solution to your equation is:
x = 1