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Solve. Check for extraneous solutions

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Hi Tracy,

So extraneous solutions are solutions that you get by solving an equation, but after you try plugging it back into your original equation, the solution does not work.

So you have the following equation:

√(7x+6) - √(9+4x) = 0

Since you have two different groups under radical signs, you want to move one to the other side so it is easier to cancel things out.

√(7x+6) - √(9+4x)     =      0

               +√(9+4x)           +√(9+4x)

Adding √(9+4x) to both sides helps cancel it out on the left side and add it to the right side:

√(7x+6) = √(9+4x)

Now you have two equations in terms of x equal to each other. Since they both have radicals, you need to do something to get rid of the radicals so you can solve the equation like you normally would.

Radicals (or square roots) can be cancelled out by squaring them. (example: (√x)2 = x ) The square root or radical symbol and the square cancel each other out, and you are left with what was under the radical sign.

Now apply that to your equation:

√(7x+6) = √(9+4x)

(√(7x+6))2 = (√(9+4x))2

 

7x+6 = 9+4x

Now you see that the radical sign has been cancelled out on both sides by squaring it. We can continue and solve the equation for x.

You want the x terms on the same side, so we'll move the 4x to the side of the 7x:

7x+6 = 9+4x

-4x           -4x

7x-4x+6 = 9

3x+6 = 9

Now we need to bring the 6 to the side where the 9 is:

3x+6 = 9

    -6      -6

3x = 9-6

3x = 3

Our last step is to get the x by itself on the left side by dividing both sides by 3:

3x    =  3

÷3      ÷3

(3x)÷3 = 3÷3

x = 1

Now we have our solutions of x=1, and we just need to check whether it is extraneous or not. We can do this by taking x=1 and plugging it back into your original equation √(7x+6) - √(9+4x) = 0.

√(7x+6) - √(9+4x) = 0

√(7(1)+6) - √(9+4(1)) = 0

√(7+6) - √(9+4) = 0

√13 - √13 = 0

0 = 0

We end up getting that 0 = 0 which is correct, thus there are no extraneous solutions.

 

For future reference if you solve an equation and plug the solution back in and get something like 0=1 or 2=3, then your solution is extraneous because the solution does not work.


I hope this helped! Let me know if you have any more questions!!

-Beatrix