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Question 99970: Please help. I have one last problem on my test due tonight. I know I am not giving you a lot of time but the person that usually helps me has ended up in the hospital and I am stuck on this last question.
Solve sqrt2x-1=sqrt4x-1 with another -1 outside the sqrt part.
Thanks if you have the time.
Answer by fastblue(13)   (Show Source):
You can put this solution on YOUR website! Square both sides,gives you 2x-1=((4x-1)SQRT-1)*((4x-1)SQRT-1)
gives you 2x-1= (4x-1)+1-2(4x-1)SQRT, move everything to one side
of the problem and square the following again. SQRT means the previous to the square root.
(2x-1)-(4x-1) -1 = -2(4x-1)SQRT
Divide each side by -2 before squareing again.
This gives you XSQ + 1/4 + X = 4X-1 after squaring(sq means square)
move everything to one side.
Xsq -3x +5/4 = 0
Multiply by 4, this gives you 4Xsq - 12X + 5=0
this breaks down to (2x - 1)(2x - 5)=0
so x=-1/2 and x=2/5
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