Best Answer

Multiply the expression (fraction) by (6 + √5)/(6 + √5) (which is 1 and will not change its value).

3/(6 - √5) x (6 + √5)/(6 + √5)

= 3(6 + √5)/(6 - √5)(6 + √5)

= 3(6 + √5)/(36-5) = 3(6 + √5)/31

= 3/31(6 + √5)

Remember:

(a + b)(a - b) = a2 - ab + ab - b2

= a2 - b2

So if either a or b contains a square root, a2 and b2 will not, and so a2 - b2 will not!

Q: How do you rationalize the denominator in this expression 3 over quantity of 6 minus square root of 5?

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1.5

The idea is to get rid of the square root in the denominator. For this purpose, you must multiply numerator and denominator by the square root of 6 in this case.

6

Square root is a mathematical function whose argument needs to be a mathematical quantity or expression. Since helium is neither a mathematical quantity nor an expression, the question makes no sense.

An example may help. If you have the fraction 1 / (2 + root(3)), where root() is the square root function, you multiply top and bottom by (2 - root(3)). If you multiply everything out, you will have no square root in the denominator, instead, you will have a square root in the numerator. If the denominator is only a root, eg root(3), you multiply top and bottom by root(3).

Related questions

1.5

Multiply everything by the square root of 3 minus the square root of 2 and then times that by 100 - 72 and divide that by 5

-26

The idea is to get rid of the square root in the denominator. For this purpose, you must multiply numerator and denominator by the square root of 6 in this case.

6

Yes. For example, the conjugate of (square root of 2 + square root of 3) is (square root of 2 - square root of 3).

Yes. The original denominator and its conjugate will form the factors of a Difference of Two Squares (DOTS) and that will rationalise the denominator but only if the radicals are SQUARE roots.

If you want to rationalize the denominator, then multiply numerator & denominator by sqrt(5), so 8*sqrt(5)/5 = approx 3.578

0.4

Square root is a mathematical function whose argument needs to be a mathematical quantity or expression. Since helium is neither a mathematical quantity nor an expression, the question makes no sense.

An example may help. If you have the fraction 1 / (2 + root(3)), where root() is the square root function, you multiply top and bottom by (2 - root(3)). If you multiply everything out, you will have no square root in the denominator, instead, you will have a square root in the numerator. If the denominator is only a root, eg root(3), you multiply top and bottom by root(3).

Depends on the situation. You usually have to multiply numerator and denominator by some number or expression. Examples: 1 / square root of 2 Here, you have to multiply numerator and denominator by the square root of 2. 1 / (square root of 2 + square root of 3) Here, you have to multiply numerator and denominator by (square root of 2 - square root of 3).