4/((square root of 7)+(square root of 3))
How do you get rid of the roots at the denominator so that it is at the simplest form?
2007-01-22 04:47:20 · 4 answers · asked by -WANTED- 3 in Science & Mathematics ➔ Mathematics
Multiply by the conjugate, sqrt[7]-sqrt[3], in both the denominator and numerator.
That multiplies to (4sqrt(7)-4sqrt(3))/(7-3)
=
(4sqrt(7)-4sqrt(3))/(4)
=
sqrt(7)-sqrt(3)
2007-01-22 04:55:44 · answer #1 · answered by Mr. Chemistry 2 · 1⤊ 0⤋
I think this requires two multiplications.
The first time, multiply top and bottom by [(square root of 7)-(square root of 3)].
The bottom becomes [4 - 2(square root of 21)].
Next multiply top and bottom by [4 + 2(square root of 21)].
2007-01-22 12:55:07 · answer #2 · answered by fcas80 7 · 1⤊ 0⤋
multiply the botton of the fraction with (sqrt(7) - sqrt(3))
it's like multiplying binomials. You will have to get rid of the sqrt's in the bottom..that is your goal.
I wish i could write it out for you, but i simply dont have the time. plus. i think you'd learn much more if you actually got HINTS and not the answer. Math is best understood by doing it on your own.
2007-01-22 12:53:20 · answer #3 · answered by roya67 2 · 1⤊ 0⤋
multiply the fraction (both top and bottom) by (sqrt7 - sqrt3)
then simpity..... my final answer is (sqrt7 - sqrt3)
2007-01-22 13:01:14 · answer #4 · answered by sh 1 · 1⤊ 0⤋