Multiplying Radicals with FOIL Method

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SUMMARY

The forum discussion focuses on the application of the FOIL method for multiplying radicals, specifically the expressions (-7 + √{3x})(4 + √{3x}) and (2 + √{5})(-2 + √{5k}). Participants clarify the correct simplification of terms, particularly the product of radicals, confirming that √{3x} * √{3x} equals 3x. The final simplified forms of the expressions are -28 - 3√{3x} + 3x and -4 + 2√{5k} - 2√{5} + 5√{k}, respectively. The discussion emphasizes the importance of clear notation and combining like terms.

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zelda1850
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tried to solve the problem with foil method but i got confuse can someone help

1) (-7 + \sqrt{3x})( 4+ \sqrt{3x})

-7 * 4 = -28
-7 * \sqrt{3x} = -7\sqrt{3x}

\sqrt{3x} * 4 = 4\sqrt{3x}
\sqrt{3x} * \sqrt{3x} - does this equal \sqrt{3x} square or is it just 3x or leave it \sqrt{3x}

this is what i got so far

= -28 + -7\sqrt{3x} + 4\sqrt{3x} + \sqrt{3x}

-7\sqrt{3x} + 4\sqrt{3x} = 3\sqrt{3x}

= -28 + 3\sqrt{3x} + \sqrt{3x}
 
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zelda1850 said:
tried to solve the problem with foil method but i got confuse can someone help

1) (-7 + \sqrt{3x})( 4+ \sqrt{3x})

-7 * 4 = -28
-7 * \sqrt{3x} = -7\sqrt{3x}

\sqrt{3x} * 4 = 4\sqrt{3x}
\sqrt{3x} * \sqrt{3x} - does this equal \sqrt{3x} square or is it just 3x or leave it \sqrt{3x}
(\sqrt{3x})(\sqrt{3x})= (\sqrt{3x})^2= 3x

this is what i got so far

= -28 + -7\sqrt{3x} + 4\sqrt{3x} + \sqrt{3x}

-7\sqrt{3x} + 4\sqrt{3x} = 3\sqrt{3x}
No, -7\sqrt{3x}+ 4\sqrt{3x}= -3\sqrt{3x}

= -28 + 3\sqrt{3x} + \sqrt{3x}
-28- 3\sqrt{3x}+ 3x

By the way, I think you will find that formulas look better if you put tex tags around the entire formula, not just "bits".
 
ok can someone check if I am doing these questions correctly now?

1) (\sqrt{2a} -5)(7\sqrt{2a} - 5)

\sqrt{2a} * 7\sqrt{2a} = 7 * 2a = 14a
\sqrt{2a} * 5 = 5\sqrt{2a}

5 * 7\sqrt{2a} = 35\sqrt{2a}
5 * 5 = 25

= 35\sqrt{2a} - 5\sqrt{2a} = 30\sqrt{2a}

= 14a + 30\sqrt{2a} - 25


2) (2 + \sqrt{5})( -2 + \sqrt{5k})

2 * -2 = -4
2 * \sqrt{5k} = 2\sqrt{5k}

\sqrt{5} * -2 = -2\sqrt{5}
\sqrt{5} * \sqrt{5k} = 25k

= -4 + 2\sqrt{5k} - -2\sqrt{5} + 25k
 
You could use one or two more steps in #2. Possibly combine like terms of square root of 5.

-4 + 2\sqrt{5k}-2\sqrt{5}+25k
-4 + (\sqrt{k}-1)2\sqrt{5} + 25k
 
im not sure what the like terms would be
 
zelda1850 said:
ok can someone check if I am doing these questions correctly now?
2) (2 + \sqrt{5})( -2 + \sqrt{5k})
Are you sure there isn't a typo in the first radical? I think the problem might actually be this:
(2 + \sqrt{5k})(-2 + \sqrt{5k})
zelda1850 said:
2 * -2 = -4
2 * \sqrt{5k} = 2\sqrt{5k}
Don't put this stuff in as separate steps. We all know that 2 * (-2) = -4 and there is absolutely no advantage in writing 2 *\sqrt{5k} = 2\sqrt{5k}.

There should be an unbroken line of expressions connected by =, ending when you have reached the simplest form of your beginning expression. Putting in these baby steps makes it more difficult to follow your work, because it isn't clear what the original expression is actually equal to.
zelda1850 said:
\sqrt{5} * -2 = -2\sqrt{5}
\sqrt{5} * \sqrt{5k} = 25k

= -4 + 2\sqrt{5k} - -2\sqrt{5} + 25k
 
that was the right question so can someone tell me what i did wrong?
 
What you did wrong is when you said \sqrt{5}\sqrt{5k}=25k.
 
oh how can i correct it would it be sqaure root 25k?
 
  • #10
Yes, that would be correct: \sqrt{5}\sqrt{5k}=\sqrt{25k}=5\sqrt{k}
 
  • #11
can i add 2\sqrt{5k} + 5\sqrt{k} or it doesn't work
 
  • #12
No, you can't add that.
 
  • #13
so my answer would be

-4 + 2\sqrt{5k} - 2\sqrt{5} + 5\sqrt{k}
 
  • #14
That would be correct :smile:
 

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