'nother eqn w/ radical: Solve 2 sqrt(x+2) - 1 = 3sqrt(x)
- Thread starter sfin
- Start date
Re: 'nother equation with radical
(3sqrt(x) + 1)^2
= (3sqrt(x) + 1) times (3sqrt(x) + 1)
= 9x + 6sqrt(x) + 1
Remember that (a + b)^2 = a^2 + 2ab + b^2
So you ain't finished: go square again (as in your other problem) :idea:
4(x+2) is correct, but not 3(x+1)sfin said:
(3sqrt(x) + 1)^2
= (3sqrt(x) + 1) times (3sqrt(x) + 1)
= 9x + 6sqrt(x) + 1
Remember that (a + b)^2 = a^2 + 2ab + b^2
So you ain't finished: go square again (as in your other problem) :idea:
Re: 'nother equation with radical
Hello, sfin!
It's hard to see your work from here . . .
We have: \(\displaystyle \:2\sqrt{x\,+\,2}\:=\:3\sqrt{x}\,+\,1\)
Square both sides carefully:
. . \(\displaystyle (2\sqrt{x\,+\,2})^2\:=\
3\sqrt{x}\,+\,1)^2\;\;\Rightarrow\;\;4(x\,+\,2)\:=\:9x\,+\,6\sqrt{x}\,+\,1\)
. . \(\displaystyle 4x\,+\,8\:=\:9x\,+\,6\sqrt{x}\,+\,1\;\;\Rightarrow\;\;-5x\,+\,7\:=\:6\sqrt{x}\)
Square both sides:
. . \(\displaystyle \-5x\,+\,7)^2\:=\6\sqrt{x})^2\;\;\Rightarrow\;\;25x^2\,-\,70x\,+\,49\:=\:36x\)
And we have a quadratic: \(\displaystyle \:25x^2\,-\,106x\,+\,49\:=\:0\)
Quadratic Formula: \(\displaystyle \:x\:=\:\frac{106\,\pm\,\sqrt{106^2\,-\,4(25)(49)}}{2(25)} \:=\:\frac{106\,\pm\,\sqrt{6336}}{50}\:=\:\frac{106\,\pm\,24\sqrt{11}}{50}\)
. . Hence: \(\displaystyle \:x\;=\;\frac{53\,\pm\,12\sqrt{11}}{25}\)
The larger root (approx. 3.712) is extraneous.
The only solution is: \(\displaystyle \L\:x\:=\:\frac{53\,-\,12\sqrt{11}}{25} \:\approx\:0.528\)
Hello, sfin!
It's hard to see your work from here . . .
We have: \(\displaystyle \:2\sqrt{x\,+\,2}\:=\:3\sqrt{x}\,+\,1\)
Square both sides carefully:
. . \(\displaystyle (2\sqrt{x\,+\,2})^2\:=\
3\sqrt{x}\,+\,1)^2\;\;\Rightarrow\;\;4(x\,+\,2)\:=\:9x\,+\,6\sqrt{x}\,+\,1\). . \(\displaystyle 4x\,+\,8\:=\:9x\,+\,6\sqrt{x}\,+\,1\;\;\Rightarrow\;\;-5x\,+\,7\:=\:6\sqrt{x}\)
Square both sides:
. . \(\displaystyle \
-5x\,+\,7)^2\:=\6\sqrt{x})^2\;\;\Rightarrow\;\;25x^2\,-\,70x\,+\,49\:=\:36x\)And we have a quadratic: \(\displaystyle \:25x^2\,-\,106x\,+\,49\:=\:0\)
Quadratic Formula: \(\displaystyle \:x\:=\:\frac{106\,\pm\,\sqrt{106^2\,-\,4(25)(49)}}{2(25)} \:=\:\frac{106\,\pm\,\sqrt{6336}}{50}\:=\:\frac{106\,\pm\,24\sqrt{11}}{50}\)
. . Hence: \(\displaystyle \:x\;=\;\frac{53\,\pm\,12\sqrt{11}}{25}\)
The larger root (approx. 3.712) is extraneous.
The only solution is: \(\displaystyle \L\:x\:=\:\frac{53\,-\,12\sqrt{11}}{25} \:\approx\:0.528\)
'nother equation with radical - SOLVED
I should refrain from doing math late at night. Things are alot clearer today. AXRW and Denis are both correct. I copied the equation from the text wrong, so the answers given (x=7, x=-1) are correct. Equation should have been:
2 sqrt(x+2) - sqrt(3x+4) = 1
2 sqrt(x+2) = sqrt(3x+4) + 1
4(x+2) = (sqrt(3x+4))^2 + 2sqrt(3x+4) + 1
4x+8 = 3x + 4 + 2sqrt(3x+4) + 1
x+3 = 2sqrt(3x+4)
x^2 + 6x + 9 = 12x + 16
x^2 - 6x - 7 = 0
(x-7)(x-1)
Both x=7 and x=-1 check out correctly. I was positive that my text answers were wrong and was believing what Denis wrote. In the end it is all my mistakes. Thanks for bearing with me on this one guys! No more late night math for me! Was a waste of my time and yours!!
Thanks
I should refrain from doing math late at night. Things are alot clearer today. AXRW and Denis are both correct. I copied the equation from the text wrong, so the answers given (x=7, x=-1) are correct. Equation should have been:
2 sqrt(x+2) - sqrt(3x+4) = 1
2 sqrt(x+2) = sqrt(3x+4) + 1
4(x+2) = (sqrt(3x+4))^2 + 2sqrt(3x+4) + 1
4x+8 = 3x + 4 + 2sqrt(3x+4) + 1
x+3 = 2sqrt(3x+4)
x^2 + 6x + 9 = 12x + 16
x^2 - 6x - 7 = 0
(x-7)(x-1)
Both x=7 and x=-1 check out correctly. I was positive that my text answers were wrong and was believing what Denis wrote. In the end it is all my mistakes. Thanks for bearing with me on this one guys! No more late night math for me! Was a waste of my time and yours!!
Thanks