# Surds

• Jun 26th 2012, 09:18 AM
GrigOrig99
Surds
I'm having a bit of trouble with this one. Can anyone help me out?

Many thanks.

Q. Given that $(a+\sqrt{3})(b-\sqrt{3})=7+3\sqrt{3}$ & that a & b are positive integers, find the value of a & b.

Attempt: $ab-a\sqrt{3}+b\sqrt{3}=7+3\sqrt{3}$ => $ab-a\sqrt{3}=3\sqrt{3}-b\sqrt{3}+10$ => $a(b-\sqrt{3})=3\sqrt{3}-b\sqrt{3}+10$ => $a=\frac{3\sqrt{3}-b\sqrt{3}+10}{b-\sqrt{3}}$ => $a=\frac{\sqrt{3}(3-b)+10}{b-\sqrt{3}}$ => $a=\frac{\sqrt{3}(3-b)+10}{b-\sqrt{3}}.\frac{b+\sqrt{3}}{b+\sqrt{3}}$ => $a=\frac{3b\sqrt{3}-b^2\sqrt{3}+10b+9-3b+10\sqrt{3}}{b^2-3}$...

Ans.: (From text book): a = 2, b = 5
• Jun 26th 2012, 09:22 AM
Prove It
Re: Surds
Quote:

Originally Posted by GrigOrig99
I'm having a bit of trouble with this one. Can anyone help me out?

Many thanks.

Q. Given that $(a+\sqrt{3})(b-\sqrt{3})=7+3\sqrt{3}$ & that a & b are positive integers, find the value of a & b.

Attempt: $ab-a\sqrt{3}+b\sqrt{3}=7+3\sqrt{3}$ => $ab-a\sqrt{3}=3\sqrt{3}-b\sqrt{3}+10$ => $a(b-\sqrt{3})=3\sqrt{3}-b\sqrt{3}+10$ => $a=\frac{3\sqrt{3}-b\sqrt{3}+10}{b-\sqrt{3}}$ => $a=\frac{\sqrt{3}(3-b)+10}{b-\sqrt{3}}$ => $a=\frac{\sqrt{3}(3-b)+10}{b-\sqrt{3}}.\frac{b+\sqrt{3}}{b+\sqrt{3}}$ => $a=\frac{3b\sqrt{3}-b^2\sqrt{3}+10b+9-3b+10\sqrt{3}}{b^2-3}$...

Ans.: (From text book): a = 2, b = 5

\displaystyle \begin{align*} \left(a + \sqrt{3}\right)\left(b - \sqrt{3}\right) &= 7 + 3\sqrt{3} \\ ab - a\sqrt{3} + b\sqrt{3} - 3 &= 7 + 3\sqrt{3} \\ ab - 3 + \left(b - a\right)\sqrt{3} &= 7 + 3\sqrt{3} \\ ab - 3 = 7 \textrm{ and } b - a &= 3 \\ ab = 10 \textrm{ and } b - a &= 3 \end{align*}

Now solve these equations simultaneously for a and b.
• Jun 26th 2012, 09:29 AM
Plato
Re: Surds
Quote:

Originally Posted by GrigOrig99
Q. Given that $(a+\sqrt{3})(b-\sqrt{3})=7+3\sqrt{3}$ & that a & b are positive integers, find the value of a & b. Ans.: (From text book): a = 2, b = 5

$(a+\sqrt{3})(b-\sqrt{3})=7+3\sqrt{3}$
$(ab-3)+(-a+b)\sqrt3=7+3\sqrt{3}$

So $ab-3=7~\&~-a+b=3.$

Solve for $a~\&~b.$
• Jun 26th 2012, 09:31 AM
johnnylam123
Re: Surds
From above, then we let b, -a are root of a quadric equation s.t. -ab=-10, b+(-a)=3
ie, equation is x^2-3x-10=0
(x+2)(x-5)=0

then (a=2,b=5) or (a=5,b=2)
by putting to original equation, answer is found
• Jun 26th 2012, 09:45 AM
HallsofIvy
Re: Surds
Quote:

Originally Posted by GrigOrig99
I'm having a bit of trouble with this one. Can anyone help me out?

Many thanks.

Q. Given that $(a+\sqrt{3})(b-\sqrt{3})=7+3\sqrt{3}$ & that a & b are positive integers, find the value of a & b.

Attempt: $ab-a\sqrt{3}+b\sqrt{3}=7+3\sqrt{3}$

There's an error in your first line. You have "FOI" but not "L"! $(-\sqrt{3})(\sqrt{3}= -3$
so this should be $ab- a\sqrt{3}+ b\sqrt{3}- 3= 7+ 3\sqrt{3}$
You have ab- 3= 7 and b- a= 3

Quote:

=> $ab-a\sqrt{3}=3\sqrt{3}-b\sqrt{3}+10$ => $a(b-\sqrt{3})=3\sqrt{3}-b\sqrt{3}+10$ => $a=\frac{3\sqrt{3}-b\sqrt{3}+10}{b-\sqrt{3}}$ => $a=\frac{\sqrt{3}(3-b)+10}{b-\sqrt{3}}$ => $a=\frac{\sqrt{3}(3-b)+10}{b-\sqrt{3}}.\frac{b+\sqrt{3}}{b+\sqrt{3}}$ => $a=\frac{3b\sqrt{3}-b^2\sqrt{3}+10b+9-3b+10\sqrt{3}}{b^2-3}$...

Ans.: (From text book): a = 2, b = 5
a= -5, b= -2 is also a solution.
• Jun 26th 2012, 09:53 AM
GrigOrig99
Re: Surds
Great. Thanks guys.