Question

Express 7+√5/3+√5 in the form a+b√5 where a and b are intergers

Answer 1

do you know how to rationalize the denominator ?

Answer 2

multiply the numerator and denominator by the conjugate of denominator
here, it'll be 3- \(\sqrt5\)
what do u get ?

Answer 3

What do u get when u multiply it with 7+√5

Answer 4

you mean multiply (1+√5) with (3-√5)
thats your numerator, you just need to foil to simply it.

Answer 5

Its 7√5 in the question can't multiply it with 3-√5 that's the part I get confused with

Answer 6

ok,
\(7\sqrt 5 \times 3 = 21 \sqrt 5\ \\7\sqrt5\times \sqrt5= 7\sqrt{5^2}=7\times 5=35 \\so \\ 7\sqrt5(3-\sqrt5)=21\sqrt5-35 \)
got it ?

Answer 7

Thank u man

Answer 8

welcome ^_^

Answer 9

The answer on my answer sheet is 4-√5 how do I get that

Answer 10

ok, then we were supposed to multiply (7+√5)
(and not 7√5 as you mentioned "Its 7√5 in the question")
then you actually foil :
\((7+√5)(3-√5)= 7*3 -7√5+3√5-√5√5 = 21+(-7+3)√5-5\\=16-4√5=4(4-√5)\)
and if you simplify the denominator, you'll get a 4 which cancels with the numerator's 4 and you are just left with 4-√5

Answer 11

I didn't get anything from that sorry but u have confused me a lot

Answer 12

i used this \((a+b)(c+d)=ac+ad+bc+bd\)

try to do this with (7+√5)(3-√5)

take 7 multiply it with 3 and -√5
then take √5, multiply it with 3 and -√5

Answer 13

to go over it:
\[ \frac{ 7+\sqrt{5}}{3+\sqrt{5}} \]

you multiply top and bottom by \( 3- \sqrt{5} \)
\[ \frac{ (7+\sqrt{5})( 3- \sqrt{5})}{(3+\sqrt{5})( 3- \sqrt{5})} \]

Answer 14

now you have two FOIL problems: one on the top and one in the bottom

can you do them ? (see hartn's posts on how)

Answer 15

I don't know how to do the 7+√5 one

Answer 16

to do
\[ (7+ \sqrt{5})(3- \sqrt{5} ) \]

you do 4 multiplies: the First, Outer, Inner, Last

This might help
http://www.khanacademy.org/math/algebra/polynomials/multiplying_polynomials/v/multiplying-binomials

Answer 17

OK thanks