SOLUTION: Problem 1: (√x+4) +2 = (√x + 20) I know the answer is 5, but do not understand how to work the problem. Problem 2: (5√x-2) -3 = (√19x-29) How do you

Algebra ->  Rational-functions -> SOLUTION: Problem 1: (√x+4) +2 = (√x + 20) I know the answer is 5, but do not understand how to work the problem. Problem 2: (5√x-2) -3 = (√19x-29) How do you       Log On


   

Question 348136: Problem 1: (√x+4) +2 = (√x + 20)
I know the answer is 5, but do not understand how to work the problem.

Problem 2: (5√x-2) -3 = (√19x-29)
How do you get 27 as the answer?
Found 2 solutions by stanbon, ewatrrr:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
Problem 1: (√x+4) +2 = (√(x + 20))
I know the answer is 5, but do not understand how to work the problem.
---
Square both sides to get:
(x+4) + 4(sqrt(x+4)) + 4 = x + 20
------
4sqrt(x+4) = 12
sqrt(x+4) = 3
----
Square both sides to get:
x+4 = 9
x = 5
===============
Problem 2: (5√x-2) -3 = (√19x-29)
Square both sides to get:
25(x-2) - 6*5sqrt(x-2) + 9 = 19x-29
----
25x-50 - 30sqrt(x-2) = 19x-38
-30sqrt(x-2) = 12-6x
------------------------
(5√x-2) -3 = (√19x-29)
---
x cannot be 2
----
Final Answer: x = 27
==========================
Cheers,
Stan H.
-5sqrt(x-2) = 2-x
----
Square both sides:
25(x-2) = 4 - 4x + x^2
25x-50 = 4 - 4x + x^2
x^2 -29x+54 = 0
(x-27)(x-2) = 0
---
x = 27 or x = 2
------------------------
Checking in original equation:



How do you get 27 as the answer?

Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!

Hi,
If I am understanding your question properly:
root%282%2C%28x%2B4%29%29%2B2+=root%282%2C+%28x%2B20%29%29
.
SQUARE BOTH SIDES OF THE EQUATION:
.
squaring left side gives:

.
squaring right gives (x+20)
.
Result of squaring both sides of the equation:
x%2B4+%2B+4%2Aroot%282%2C%28x%2B4%29%29+%2B+4+=+%28x+%2B+20%29
.
Simplifying
root%282%2C%28x%2B4%29%29=+3
.
SQUARING BOTH SIDES OF THIS EQUATION
x+4 = 9
x=5
.
this is solved as above: fist Squaring both sides of the equation, simplifying and once again squaring both sides of the equation to solve for x
5%2Aroot%282%2C%28x-2%29%29-3+=root%282%2C+%2819%2Ax%2B29%29%29
Square both sides to get:
25%28x-2%29+-+30sqrt%28x-2%29+%2B+9+=+19x-29
25x-50+-+30sqrt%28x-2%29+=+19x-38
-30sqrt%28x-2%29+=+12-6x
%286x-12%29=+30sqrt%28x-2%29
%28x-2%29=+5sqrt%28x-2%29
.
SQUARE BOTH SIDES
x%5E2+-4x+%2B4+=+25%28x-2%29
x%5E2+-4x+%2B4+=+25x-50%29
x%5E2+-29x+%2B54=+0
.
factor
%28x-27%29%28x-2%29=0
x=27 or x=2