MAths Advanced - Surds Help (1 Viewer)

[kaha167]

Hey everyone,
I was wonderning if anyone would happen to have an overview of the surds topic
My teacher never covered it and i only have a confusing textbook
Any help is really appreciated.
I'm desperate.
Thankyou so much

 

[lyounamu]

Hey everyone,
I was wonderning if anyone would happen to have an overview of the surds topic
My teacher never covered it and i only have a confusing textbook
Any help is really appreciated.
I'm desperate.
Thankyou so much

Here are some rules:

SQ(A) x SQ(B) = SQ(AB)

e.g.

SQ(3) x SQ(4) = SQ(12)


SQ(A^2) = A

e.g.

SQ(9) = SQ(3^2) = 3

SQ(A) divided by SQ(B) = SQ(A/B)

SQ 5 divided by SQ 6 = SQ(5/6)

 

[supercalamari]

Funnily enough, surds is THE only maths thing I am good at. I was going to post stuff, but Namu's done it for me. Oh well, ciao

 

[kaha167]

hey,
thanks for your help
but i don't understand the SQ
what does that mean?

 

[supercalamari]

hey,
thanks for your help
but i don't understand the SQ
what does that mean?

I'm assuming that SQ means 'square root of'.

 

[JasonNg1025]

Couple more tips (I'm used to sqrt(...) being square root... so... yah)

sqrt (x) = x^1/2

sqrt(a) + sqrt(b) =/= sqrt(a + b)

a + xsqrt(b) is the conjugate of a - xsqrt(b), and vice versa
(So a non-surd + surd is the conjugate of a non-surd - surd)

Just as a note, conjugate surds multiply to get a rational number.
[a + xsqrt(b)][a - xsqrt(b)]
= (a)² - (xsqrt{b})²
= a² - x²b.

From this, there's this thing called rationalising the denominator.
So say there's a fraction, and the denominator contains a surd. (NOTE - all surds should be able to be expressed in the form a + xsqrt(b) where a and b are rational).
Then, you multiply both numerator and denominator by the conjugate of the denominator, that is, a - xsqrt(b).
The result should be that you have a surd in the numerator and a rational denominator.

Example:

Rationalise the denominator of 5 / (3 + sqrt(2)) - (made up on the spot)
First, we multiply both by the conjugate. This is 3 - sqrt(2). We get:

( 15 - 5sqrt(2) ) / ( 9 - 2 ), and simplifying, we get

( 15 - 5sqrt(2) ) / 7, or, to put it in the form a + sqrt(b);

( 15 / 7 ) - ( 5 / 2 ) sqrt(2)

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Hope this helps!