Multiplying Radicals with FOIL Method

[zelda1850]

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$$\sqrt{3x}$$ + $$\sqrt{3x}$$

 

[HallsofIvy]

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".

 

[zelda1850]

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

 

[symbolipoint]

You could use one or two more steps in #2. Possibly combine like terms of square root of 5.

$$-4 + 2\sqrt{5k}-2\sqrt{5}+25k$$
$$-4 + (\sqrt{k}-1)2\sqrt{5} + 25k$$

 

[zelda1850]

im not sure what the like terms would be

 

[Mark44]

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

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.

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

= -4 + $$2\sqrt{5k}$$ - $$-2\sqrt{5}$$ + 25k

 

[zelda1850]

that was the right question so can someone tell me what i did wrong?

 

[micromass]

What you did wrong is when you said $$\sqrt{5}\sqrt{5k}=25k$$.

 

[zelda1850]

oh how can i correct it would it be sqaure root 25k?

 

[micromass]

Yes, that would be correct: $$\sqrt{5}\sqrt{5k}=\sqrt{25k}=5\sqrt{k}$$

 

[zelda1850]

can i add $$2\sqrt{5k}$$ + $$5\sqrt{k}$$ or it doesn't work

 

[micromass]

No, you can't add that.

 

[zelda1850]

so my answer would be

-4 + $$2\sqrt{5k}$$ - $$2\sqrt{5}$$ + $$5\sqrt{k}$$

 

[micromass]

That would be correct