SOLUTION: how do you solve the following: {{{sqrt(x+4)}}} + {{{sqrt(2x-1)}}} = {{{sqrt(7x+1)}}}

Algebra ->  Radicals -> SOLUTION: how do you solve the following: {{{sqrt(x+4)}}} + {{{sqrt(2x-1)}}} = {{{sqrt(7x+1)}}}       Log On


   

Question 519668: how do you solve the following:
sqrt%28x%2B4%29 + sqrt%282x-1%29 = sqrt%287x%2B1%29

Found 3 solutions by tracyc0912, Edwin McCravy, MathTherapy:
Answer by tracyc0912(40) About Me  (Show Source):
You can put this solution on YOUR website!
You need to square the whole darn thing and it removes the radical signs
1x+4+2x-1=7x+1 and simplify
Combine like terms
3x+3=7x+1 (subtract 7x from each side)
-4x+3=1 (subtract 3 from each side)
-4x=-2 (divide both sides by -4)
x=-2/4, reduces to
x=-1/2
I hope this helps! :-)

Answer by Edwin McCravy(20077) About Me  (Show Source):
You can put this solution on YOUR website!

The other tutor's solution is incorrect

sqrt%28x%2B4%29 + sqrt%282x-1%29 = sqrt%287x%2B1%29

Since one of the radicals is isolated, we square both sides:

To make things a little easier, 

let A = sqrt%28x%2B4%29
let B = sqrt%282x-1%29
let C = sqrt%287x%2B1%29

Then the above equation becomes:

         A + B = C

Squaring both sides:

      (A + B)² = C²

(A + B)(A + B) = C²

 A² + 2AB + B² = C² 

Since A = sqrt%28x%2B4%29, then A² = x+4
Since B = sqrt%282x-1%29, then B² = 2x-1
let C = sqrt%287x%2B1%29, then C² = 7x+1

So now we have

 x+4 + 2AB + 2x-1 = 7x+1

Simplify and solve for 2AB

 3x + 3 + 2AB = 7x + 1
          2AB = 4x - 2

Divide through by 2 since all coefficients are even:

           AB = 2x - 1

Square both sides again:

        (AB)² = (2x - 1)²

         A²B² = (2x - 1)(2x - 1)              

         A²B² = 4x² - 4x + 1

Now substitute A² = x+4  and B² = 2x-1 from above:

 (x+4)(2x-1) = 4x² - 4x + 1 

 2x² + 7x - 4 = 4x² - 4x + 1

 -2x² + 11x - 5 = 0

  2x² - 11x + 5 = 0

(x - 5)(2x - 1) = 0

x - 5 = 0;  2x - 1 = 0

    x = 5;      2x = 1

                 x = 1%2F2

Check each solution as they may be extraneous: 
             
Checking x = 5

sqrt%28x%2B4%29 + sqrt%282x-1%29 = sqrt%287x%2B1%29
sqrt%285%2B4%29 + sqrt%282%285%29-1%29 = sqrt%287%285%29%2B1%29
sqrt%289%29 + sqrt%289%29 = sqrt%2836%29
3 + 3 = 6
    6 = 6

That checks so x = 5 is a solution.

Checking x = 1%2F2

sqrt%28x%2B4%29 + sqrt%282x-1%29 = sqrt%287x%2B1%29
sqrt%281%2F2%2B4%29 + sqrt%282%281%2F2%29-1%29 = sqrt%287%281%2F2%29%2B1%29
sqrt%281%2F2%2B8%2F2%29 + sqrt%281-1%29 = sqrt%287%2F2%2B1%29
sqrt%289%2F2%29 + sqrt%280%29 = sqrt%287%2F2%2B2%2F2%29
sqrt%289%2F2%29 + 0 = sqrt%289%2F2%29
sqrt%289%2F2%29 = sqrt%289%2F2%29

That checks also, so x = 1%2F2 is also a solution.

Edwin

Answer by MathTherapy(10719) About Me  (Show Source):
You can put this solution on YOUR website!
how do you solve the following:

sqrt%28x%2B4%29 + sqrt%282x-1%29 = sqrt%287x%2B1%29
******************************
One of the persons who responded states that x+=+-+1%2F2, but this is WRONG!! x CANNOT equal -+1%2F2.

Of the 3 RADICAL expressions, 2x - 1 is the SMALLEST. So, 2x+-+1+%3E=+0, and 2x+%3E=+1 ===> x+%3E=+1%2F2.
  So, we now get: sqrt%28x+%2B+4%29 + sqrt%282x+-+1%29 = sqrt%287x+%2B+1%29, with x+%3E=+1%2F2
                  sqrt%28x+%2B+4%29 + sqrt%282x+-+1%29 = sqrt%287x+%2B+1%29
                  %28sqrt%28x+%2B+4%29+%2B+sqrt%282x+-+1%29%29%5E2 = %28sqrt%287x+%2B+1%29%29%5E2 ----- Squaring both sides
 = 7x + 1
     x+%2B+4+%2B+2sqrt%28%28x+%2B+4%29%282x+-+1%29%29+%2B+2x+-+1 = 7x + 1
              3x+%2B+3+%2B+2sqrt%282x%5E2+%2B+7x+-+4%29 = 7x + 1
                     2sqrt%282x%5E2+%2B+7x+-+4%29 = 7x + 1 - 3x - 3
                     2sqrt%282x%5E2+%2B+7x+-+4%29 = 4x - 2
                  %282sqrt%282x%5E2+%2B+7x+-+4%29%29%5E2 = %284x+-+2%29%5E2 ------ Squaring both sides
                       4%282x%5E2+%2B+7x+-+4%29+=+16x%5E2+-+16x+%2B+4
                         8x%5E2+%2B+28x+-+16+=+16x%5E2+-+16x+%2B+4
                                    0+=+16x%5E2+-+16x+%2B+4+-+8x%5E2+-+28x+%2B+16
                                    0+=+8x%5E2+-+44x+%2B+20
                                    0+=+4%282x%5E2+-+11x+%2B+5%29
                                    0+=+2x%5E2+-+11x+%2B+5
                                    0+=+2x%5E2+-+10x+-+x+%2B+5
                                    0+=+2x%28x+-+5%29+-+1%28x+-+5%29
                                    0 = (2x - 1)(x - 5) 
                                    0 = 2x - 1      OR     0 = x - 5
                                    1 = 2x          OR    5 = x
                                   highlight%281%2F2+=+x%29
As x+=+1%2F2+%3E=+1%2F2 and x+=+5+%3E=+1%2F2, both solutions for x are VALID!