SOLUTION: I can only solve part of this equation and then I get stuck. Can someone show me the next steps I need to take and explain how they are done? I have done the equation as far as I

Algebra ->  Square-cubic-other-roots -> SOLUTION: I can only solve part of this equation and then I get stuck. Can someone show me the next steps I need to take and explain how they are done? I have done the equation as far as I       Log On


   

Question 60914: I can only solve part of this equation and then I get stuck. Can someone show me the next steps I need to take and explain how they are done? I have done the equation as far as I can:
SQRT(2X-1)+SQRT(X-4)= 4 <----original equation
SQRT(2X-1)=4-SQRT(X-4)
(SQRT<2X-1>)^2 = (4-SQRT)^2
2X-1 = (4-SQRT)(4-SQRT) <------I am lost as to how to solve finish this side of the equation.
Thanks!
Found 2 solutions by funmath, joyofmath:
Answer by funmath(2933) About Me  (Show Source):
You can put this solution on YOUR website!
sqrt%282X-1%29%2Bsqrt%28X-4%29=4 <----original equation
sqrt%282X-1%29=4-sqrt%28X-4%29
sqrt%282X-1%29%5E2=%284-sqrt%28X-4%29%29%5E2
2X-1=%284-sqrt%28X-4%29%29%284-sqrt%28X-4%29%29
2x-1=4%284-sqrt%28x-4%29%29-sqrt%28x-4%29%284-sqrt%28x-4%29%29
2x-1=16-4sqrt%28x-4%29-4sqrt%28x-4%29%2Bsqrt%28x-4%29%5E2
2x-1=16-8sqrt%28x-4%29%2Bx-4
2x-1=-8sqrt%28x-4%29%2Bx%2B12
2x-x-1-12=-8sqrt%28x-4%29%2Bx-x%2B12-12
x-13=-8sqrt%28x-4%29
%28x-13%29%5E2=%28-8sqrt%28x-4%29%29%5E2
x%5E2-26x%2B169=64%28sqrt%28x-4%29%29%5E2
x%5E2-26x%2B169=64%28x-4%29
x%5E2-26x%2B169=64x-256
x%5E2-26x-64x%2B169%2B256=0
x%5E2-90x%2B425=0
(x-5)(x-85)=0
x-5=0 and x-85=0
x=5 and x=85
Check for extraneous solutions:
Let x=5
sqrt%282%285%29-1%29%2Bsqrt%28%285%29-4%29=4
sqrt%2810-1%29%2Bsqrt%285-4%29=4
sqrt%289%29%2Bsqrt%281%29=4
3%2B1=4
4=4 x=5 is a valid solution.
let x=85
sqrt%282%2885%29-1%29%2Bsqrt%28%2885%29-4%29=4
sqrt%28170-1%29%2Bsqrt%2885-4%29=4
sqrt%28169%29%2Bsqrt%2881%29=4
13%2B9=4
22=4 x=85 is an extraneous solution, it's not valid.
Therefore the solution is: highlight%28x=5%29
Happy Calculating!!!

Answer by joyofmath(189) About Me  (Show Source):
You can put this solution on YOUR website!
The equation to solve is sqrt%282x-1%29%2Bsqrt%28x-4%29=4.
This is a quadratic equation in disguise!
The trick to solving these is to get all the square root stuff onto one side of the equation then square both sides. Then, you're left with a quadratic equation to solve:
Square both sides of the equation to solve:
%28sqrt%282x-1%29%29%5E2+%2B+sqrt%28x-4%29%5E2+=+4%5E2+=+16
This expands to:
%282x-1%29%2B%28x-4%29%2B2sqrt%28%282x-1%29%28x-4%29%29+=+16
Collect like terms:
3x-5%2B2sqrt%28%282x-1%29%28x-4%29%29+=+16
Subtract 16 from both sides:
3x-21%2B2sqrt%28%282x-1%29%28x-4%29%29+=+0
Subtract 2sqrt%28%282x-1%29%28x-4%29%29 from both sides:
3x-21+=+-+2sqrt%28%282x-1%29%28x-4%29%29
Now, square both sides:
%283x-21%29%5E2+=+%28-2sqrt%28%282x-1%29%28x-4%29%29%29%5E2
Multiply things out:
9x%5E2-126x%2B441+=+4%282x-1%29%28x-4%29
Or 9x%5E2-126x%2B441+=+4%282x%5E2-9x%2B4%29+=+8x%5E2-36x%2B16
Subtract 8x%5E2 from both sides:
x%5E2-126x%2B441=-36x%2B16
Subtract 16 from both sides:
x%5E2-126x%2B425=-36x
Add 36x to both sides:
x%5E2-90x%2B425=0
This is a quadratic equation that you can solve with the quadratic formula or by factoring the equation if you can see the factors:
The equation factors into %28x-85%29%28x-5%29
The roots of the equation are 85 and 5
If you plug 85 into your original equation it doesn't work but 5 does so 5 is your answer.