SOLUTION: Solve each radical equation, check for extraneous solutions: x + 1 = sqrt(7x +15)

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Question 845167: Solve each radical equation, check for extraneous solutions:
x + 1 = sqrt(7x +15)
Found 2 solutions by ewatrrr, pmesler:
Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!
 

Hi,

x + 1 = sqrt(7x +15)   |Squaring both sides

 x^2 + 2x + 1 = 7x +15

 x^2 -5x - 14 = 0

(x-7)(x+2) = 0,  x = 7   

 N0te:   x = -2  is an extraneous root  -1 ±  1

Answer by pmesler(52) About Me  (Show Source):
You can put this solution on YOUR website!
The first thing you want to do is remove the square root from the equation. To do that, you simply need to square both sides. You square because squaring is the inverse function of a square root and that removes the square root function.
(x+1)^2 = (sqrt(7x+15))^2
(x+1)(x+1) = (7x+15).

Use FOIL method to find and create the expression on the left.
x^2+2x+1 = 7x+15

Subtract 7x from both sides to get

x^2 -5x + 1 = 15
Subtract 15 from both sides to get

x^2-5x-14 = 0. What we are now left with is a polynomial in the quadratic form.
The general form for a quadratic equation is ax^2+bx+c = 0.

Here a = 1 since there is no coefficient written next to x^2 we assume it is 1,
b = -5 and c = -14.

Now we simply use the quadratic formula to find x.
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B-5x%2B-14+=+0) has the following solutons:

x%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca

For these solutions to exist, the discriminant b%5E2-4ac should not be a negative number.

First, we need to compute the discriminant b%5E2-4ac: b%5E2-4ac=%28-5%29%5E2-4%2A1%2A-14=81.

Discriminant d=81 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28--5%2B-sqrt%28+81+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%28-5%29%2Bsqrt%28+81+%29%29%2F2%5C1+=+7
x%5B2%5D+=+%28-%28-5%29-sqrt%28+81+%29%29%2F2%5C1+=+-2

Quadratic expression 1x%5E2%2B-5x%2B-14 can be factored:
1x%5E2%2B-5x%2B-14+=+1%28x-7%29%2A%28x--2%29
Again, the answer is: 7, -2. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B-5%2Ax%2B-14+%29


x = 7 or x = -2.
Now to find the extraneous roots, we substitute 7 into the original equation and see if everything checks out.
x+1 = sqrt(7x+15) <--------------> 7 + 1 = sqrt(7(7)+15)


8 = sqrt(49+15)
8 = sqrt(64)
8=8. That solution checks out.
Now, let's try -2.

-2+1 = sqrt(7(-2)+15)
-1 = sqrt(-14 + 15)
-1 = sqrt (1)
-1 = 1. This is obviously false, so -2 is the extraneous root. X = 7 is the correct root.