## blackjesus619 Group Title Express 7+√5/3+√5 in the form a+b√5 where a and b are intergers one year ago one year ago

1. hartnn

do you know how to rationalize the denominator ?

2. hartnn

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

3. blackjesus619

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

4. hartnn

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

5. blackjesus619

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

6. hartnn

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 ?

7. blackjesus619

Thank u man

8. hartnn

welcome ^_^

9. blackjesus619

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

10. hartnn

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

11. blackjesus619

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

12. hartnn

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

13. phi

to go over it: $\frac{ 7+\sqrt{5}}{3+\sqrt{5}}$ you multiply top and bottom by $$3- \sqrt{5}$$ $\frac{ (7+\sqrt{5})( 3- \sqrt{5})}{(3+\sqrt{5})( 3- \sqrt{5})}$

14. phi

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)

15. blackjesus619

I don't know how to do the 7+√5 one

16. phi

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

17. blackjesus619

OK thanks