Also:
Solve the equation for solution and check for extraneous solutions:
√y+3 = 2√y-7
2006-07-26 10:31:54 · 6 answers · asked by Brandon ツ 3 in Science & Mathematics ➔ Mathematics
6 / (3 - sqrt(3) ) ----> use the conjugate of "3 - sqrt(3)", which is
"3 + sqrt(3)"
Multiply the conjugate with the numerator AND the denominator to keep the fraction balance (and the same value):
[6 * (3 + sqrt(3)) ] / [(3 - sqrt(3)) * (3 + sqrt(3))]
=[6 * (3 + sqrt(3)) ] / (9 - 3sqrt(3) + 3sqrt(3) - 3)
=[6 * (3 + sqrt(3)) ] / (6)
= 3 + sqrt(3)
The denominator is now "1" and it is rationalized because there is no "sqrt" in it. Using the conjugate allowed you to do this.
2006-07-26 14:07:52 · answer #1 · answered by Anonymous · 1⤊ 0⤋
Using the property that (x-y)(x+y) = x^2 - y^2, multiply the fraction by (3 + √3)/(3 + √3):
6*(3 + √3)/[(3 - √3)*(3 + √3)]
= 6*(3 + √3)/[3^2 - (√3)^2]
= 6*(3 + √3)/(9 - 3)
= 6*(3 + √3)/6
= 3 + √3
Ta da!
2006-07-26 10:51:15 · answer #2 · answered by Ѕємι~Мαđ ŠçїєŋŧιѕТ 6 · 0⤊ 0⤋
y=100
2006-07-27 06:41:06 · answer #3 · answered by seba 1 · 0⤊ 0⤋
1. 3 + radical 3
2. y = 100
2006-07-27 10:32:07 · answer #4 · answered by MollyMAM 6 · 0⤊ 0⤋
Ok...I was just flipping through some of your past posts...and they are all homework problems!
What do you have your whole homework on here?
There's no point in getting an A on the homework if you fail the exam.
2006-07-26 10:36:01 · answer #5 · answered by Heather 4 · 0⤊ 0⤋
1. 3+SQRT(3)
2. y = 10
2006-07-26 15:13:58 · answer #6 · answered by jogimo2 3 · 0⤊ 0⤋