# Help with Surds

• Oct 2nd 2006, 10:03 AM
Yppolitia
Help with Surds
Hi, I basically need some help with Surds and understand them and also general algebra. Basically the first question of mine is what is about simplifying:

(Ill replace Square Root icon with a '*' as I cant find the icon anywhere to copy and paste)

1. *1 + *2 + *4 + *8

I originally thought that it would be just adding the numbers together to get *15 but then realised that isnt possible. Could someone please explain what exactly I have to do, not just give me the answer so I know how to do it in the future.

Also what would x(squared)+x-2 be? I know it would start as:

(x+ )(x- ) but dont know what goes in the gaps?

Thanks a lot.
• Oct 2nd 2006, 10:32 AM
topsquark
Quote:

Originally Posted by Yppolitia
(Ill replace Square Root icon with a '*' as I cant find the icon anywhere to copy and paste)

*1 + *2 + *4 + *8

This is really two separate questions, so I'll answer them that way. I'm also going to rewrite your problem as:
sqrt(1) + sqrt(2) + sqrt(4) + sqrt(8)

Now, we know sqrt(1) and sqrt(4) are 1 and 2 respectively. And sqrt(2) is, well, sqrt(2). We can't really do anything about this one. So we've got:
sqrt(1) + sqrt(2) + sqrt(4) + sqrt(8) = 1 + sqrt(2) + 2 + sqrt(8) = 3 + sqrt(2) + sqrt(8)

What about sqrt(8)? Well, 8 = 2*4 = 2*(2^2). So
sqrt(8) = sqrt[2*(2^2)] = sqrt(2)*sqrt[(2^2)]

But sqrt[(2^2)] = 2 since the square root and the ^2 undo each other. (Or you could simply say sqrt[(2^2)] = sqrt(4) = 2.) So
sqrt(8) = sqrt[2*(2^2)] = sqrt(2)*sqrt[(2^2)] = sqrt(2)*2 = 2*sqrt(2)

So
sqrt(1) + sqrt(2) + sqrt(4) + sqrt(8) = 3 + sqrt(2) + sqrt(8) = 3 + sqrt(2) + 2*sqrt(2)
= 3 + 3*sqrt(2)

-Dan
• Oct 2nd 2006, 10:47 AM
Soroban
Hello, Yppolitia!

Quote:

I basically need some help with Surds and understand them.
. . . . . . . . . _ . . ._ . . ._ . . _
1) Simplify: √1 + √2 + √4 + √8

. . . . . . . . . . . . _ . . . . . _ . . . . . _ . . . ._
We know that: .√1 = 1, . √4 = 2, . √8 = 2√2
. . . . . . . . . . . . . _ . . . . . . _ . . . . . . . . _
So we have: .1 + √2 + 2 + 2√2 . = . 3 + 3√2

Quote:

Also: . x² + x - 2 .= .(x + .)(x - .)

We know that the product of the two numbers is -2.
There are only two choices: 1 and -2 . . . 2 and -1

Try both and see which one give us the original trinomial.

. . (x + 1)(x - 2) .= .x² - 2x + x - 2 .= .- x - 2 . . . no

. . (x + 2)(x - 1) .= .x² - x + 2x - 2 .= .x² + x - 2 . . . yes!

• Oct 2nd 2006, 10:49 AM
topsquark
Quote:

Originally Posted by Yppolitia
Also what would x(squared)+x-2 be? I know it would start as:

(x+ )(x- ) but dont know what goes in the gaps?

I'm going to do a different problem, then get to yours. The reason is that the method I'm going to show you is better demonstrated on a more complicated problem. This method is called the "ac method."

Factor: x^2 - x - 6

What we wish to do is first list all the factors of -6:
-6, 1
-3, 2
-2, 3
-1, 6
1, -6
2, -3
3, -2
6, -1

Now, which of these combinations add to form the coefficient of the linear term? (Which is -1 in this case.) The answer is that there are two pairs listed that sums to -1: -3, 2 and 2, -3. It doesn't matter which one we pick, so I'll arbitrarily pick the first one.

So what I'm going to do is rewrite the linear term in the original problem as -x = -3x + 2x:
x^2 - x - 6 = x^2 - 3x + 2x - 6

Now I'm going to use the method "factoring by grouping."
x^2 - 3x + 2x - 6
(x^2 - 3x) + (2x - 6)
x(x - 3) + 2(x - 3)
(x + 2)(x - 3).

I'll leave it to you to check that if we had picked the 2, -3 pair that we get the same answer.

The only thing to add if we have the more general problem of factoring
ax^2 + bx + c
is that instead of factoring "c" like we did above, we are going to factor a*c. (Hence the name "ac method.") The rest is identical with what I've shown you above.

x^2+x-2

List all the factors of 1*-2 = -2 (a = 1, c = -2 here):
-2, 1
-1, 2
1, -2
2, -1
Which of these pairs add up to the coefficient of the linear term in the problem, which is 1 in this case? The answer is both -1, 2 and 2, -1 do. I'll again arbitrarily pick the first pair. Again it doesn't matter which one we pick.

So I'm going to write x = -x + 2x in the original.
x^2 + x - 2
x^2 - x + 2x - 2
(x^2 - x) + (2x - 2)
x(x - 1) + 2(x - 1)
(x + 2)(x - 1).

Now, I'll admit that in this case the answer would have been easily gotten to by guessing:
x^2 + x - 2 = (x + _)(x + _)
The only possible numbers that can go in the slots are the factors of -2 as we listed earlier, so there are only 4 possibilities to try out.

-Dan
• Oct 2nd 2006, 10:56 AM
Yppolitia
Great, thanks a lot. Sorry to be greedy but if you could help me answer these two other questions that would be very helpful :)

What would x(squared)+x-2 be? I know it would start as:

(x+ )(x- ) but dont know what goes in the gaps?

And I need to reduce the following root to the form a(sqrt)b where 'a' is an integer and 'b' in as small as possible:

(sqrt)50 = 2(sqrt)25 = 10(sqrt)5 (Is that right?)

If that is right how would I do (sqrt)99 and (sqrt)1000.

Really sorry for all these questions but my teachers is pretty arrogant and doesnt seem to like helping me.
• Oct 2nd 2006, 11:42 AM
topsquark
Quote:

Originally Posted by Yppolitia
What would x(squared)+x-2 be? I know it would start as:

(x+ )(x- ) but dont know what goes in the gaps?

Ummm...both Soroban and I went over this. Are you saying you are still having problems with it?

Quote:

Originally Posted by Yppolitia
And I need to reduce the following root to the form a(sqrt)b where 'a' is an integer and 'b' in as small as possible:

(sqrt)50 = 2(sqrt)25 = 10(sqrt)5 (Is that right?)

Yes, that is perfect! :)

Quote:

Originally Posted by Yppolitia
If that is right how would I do (sqrt)99 and (sqrt)1000.

sqrt(99) = sqrt(9*11) = sqrt(9)*sqrt(11) = 3*sqrt(11)

sqrt(1000) = sqrt(100*10) = sqrt(100)*sqrt(10) = 10*sqrt(10)

-Dan
• Oct 2nd 2006, 11:42 PM
earboth
Quote:

Originally Posted by Yppolitia
...
And I need to reduce the following root to the form a(sqrt)b where 'a' is an integer and 'b' in as small as possible:
(sqrt)50 = 2(sqrt)25 = 10(sqrt)5 (Is that right?)
...

Hi,

if I understand your problem right, you want to calculate partially the square-root of 50 (?):

sqrt(50) = sqrt(25 * 2) = sqrt(25) * sqrt(2) = 5*sqrt(2)

so, if I understand you right, your result is not right. (so sorry)

tschüss

EB