K Kaelee New member Joined Dec 3, 2006 Messages 1 Dec 3, 2006 #1 The problem is: . . .square root of the fraction 2/5x According to the book, the answer is square root of 50x over 5x. How did they get that?
The problem is: . . .square root of the fraction 2/5x According to the book, the answer is square root of 50x over 5x. How did they get that?
M mts New member Joined Dec 1, 2006 Messages 11 Dec 3, 2006 #2 Don't know. Rationalise the denominator sqrt(2)/sqrt(5x) Multiply top and bottom by sqrt(5x) (sqrt(5x)(sqrt(2))/(sqrt(5x)(sqrt(5x)) sqrt(10x)/(5x)
Don't know. Rationalise the denominator sqrt(2)/sqrt(5x) Multiply top and bottom by sqrt(5x) (sqrt(5x)(sqrt(2))/(sqrt(5x)(sqrt(5x)) sqrt(10x)/(5x)
stapel Super Moderator Staff member Joined Feb 4, 2004 Messages 16,579 Dec 3, 2006 #3 Kaelee said: square root of the fraction 2/5x Click to expand... Do you mean any of the following? . . . . .sqrt[2] / (5x) . . . . .sqrt[2/5] x . . . . .sqrt[2/(5x)] Thank you. Eliz.
Kaelee said: square root of the fraction 2/5x Click to expand... Do you mean any of the following? . . . . .sqrt[2] / (5x) . . . . .sqrt[2/5] x . . . . .sqrt[2/(5x)] Thank you. Eliz.
