Need help with Surds men (1 Viewer)

youngsky · Feb 9, 2013

include working out please

thanks guys

asianese · Feb 9, 2013

In Q7: multiply the first fraction by the conjugate on the numerator and denominator. Do the same with 3/root3 = root3. Cross multiply.

Q8: multiply the LHS by the conjugate of the denominator. Solve.

Q9: expand.

SpiralFlex · Feb 9, 2013

1.

(Slow and steady, since I believe you are only beginning this topic)

$\frac{1}{2+\sqrt{3}}+\frac{3}{\sqrt{3}}$

(For simplicity simplify the second term)

$=\frac{1}{2+\sqrt{3}}+\frac{3\sqrt{3}}{3}$

$=\frac{1}{2+\sqrt{3}}+\sqrt{3}$

Rationalise the first expression by multiplying it with its conjugate, ie. 2-sqrt(3). We do this so we have a difference between two squares - eliminating the surd expression on the denominator.

$=\frac{1}{2+\sqrt{3}}*\frac{2-\sqrt{3}}{2-\sqrt{3}}+\sqrt{3}$

$=\frac{2-\sqrt{3}}{2^2-(\sqrt{3})^2}+\sqrt{3}$

$=\frac{2-\sqrt{3}}{4-3}+\sqrt{3}$

$=2-\sqrt{3}+\sqrt{3}$

$=2$

The number two belongs to the set of rational numbers. (A number that can be expressed in the quotient form p/q where q is not 0.)

2.

$=\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}*\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}}$

Rationalise it by multiplying by the conjugate of the denominator

$=\frac{(\sqrt{3}-\sqrt{2})^2}{\sqrt{3}^2-\sqrt{2}^2}$

$=5-\sqrt{6}$

$\equiv x-y\sqrt{6}$

By the equality of surds,

$x=5,y=1$

3. See if you can use a similiar technique to do the last question. Start by expanding the left hand side.

asianese · Feb 9, 2013

kev- · Feb 9, 2013

Spiral, second line of your working out is wrong. It's supposed to be 3rt3/3 instead of 3rt3/rt3

asianese · Feb 9, 2013

Just a typo...

kev- · Feb 9, 2013

yeah sorry. Just pointed out the typo in case it confuses anyone.

youngsky · Feb 12, 2013

Thanks for the help!

@Spiral: For question 8 (the second one) when I multiplied by the conjugate I got the answer x = 5, y = 2. I think you forgot to include the other negative root 6

@asianese: Can you explain the equating the rational/irrational parts method? Don't think I've done this before.

SpiralFlex · Feb 12, 2013

Very well spotted. A mistake on my part. Thank you for correcting me.

$(\sqrt{3}-\sqrt{2})^2$

Is indeed

$3-2\sqrt{6}+2$

$5-2\sqrt{6}$

$x=5,y=-2$

SpiralFlex · Feb 12, 2013

From Asianese's line:

$a^2+2\sqrt{2}ab+2b^2=3+\sqrt{2}$

Notice on the RHS, we have something in the form

$x+y\sqrt{2}$

We can arrange the LHS into the form,

$a^2+2b^2+2\sqrt{2}ab=3+2\sqrt{2}$

And by the equality of surds. (Just like equating coefficients for polynomials) We have our two equations of -

$a^2+2b^2=3$

$2ab=2$

SpiralFlex · Feb 12, 2013

kev- said:
Spiral, second line of your working out is wrong. It's supposed to be 3rt3/3 instead of 3rt3/rt3

Thank you for pointing it out Sir. Well spotted.

Need help with Surds men (1 Viewer)

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