Simplifying Radical Expressions

[DanthePolishman]

Hey guys,

I understand the problem 2/1 - sqrt(3). You just multiply by 1 + sqrt(3). What if the denominator has a square root that has a coefficient such as sqrt(13)/2*sqrt(7)? Do you just first divide by 2 and then subtract sqrt(7)? If that, then this is the worked out solution. sqrt(13)/2*sqrt(7) > [sqrt(13)/2]/sqrt(7) > {* sqrt(-7)} > [sqrt(-91)/2]/-1 > {* (-1)} > sqrt(91)/2. Would this be correct?

Thanks,
Dan.

 

[pka]

I understand the problem 2/1 - sqrt(3). You just multiply by 1 + sqrt(3). What if the denominator has a square root that has a coefficient such as sqrt(13)/2*sqrt(7)? Do you just first divide by 2 and then subtract sqrt(7)? If that, then this is the worked out solution. sqrt(13)/2*sqrt(7) > [sqrt(13)/2]/sqrt(7) > {* sqrt(-7)} > [sqrt(-91)/2]/-1 > {* (-1)} > sqrt(91)/2. Would this be correct?

\(\displaystyle \dfrac{a}{{1 + \sqrt b }} = \dfrac{{a(1 - \sqrt b) }}{{1 - b}}\) & \(\displaystyle \dfrac{c}{{\sqrt d }} = \dfrac{{c\sqrt d }}{d}\)

 

[JeffM]

Hey guys,

I understand the problem 2/1 - sqrt(3). You just multiply by 1 + sqrt(3). What if the denominator has a square root that has a coefficient such as sqrt(13)/2*sqrt(7)? Do you just first divide by 2 and then subtract sqrt(7)? If that, then this is the worked out solution. sqrt(13)/2*sqrt(7) > [sqrt(13)/2]/sqrt(7) > {* sqrt(-7)} > [sqrt(-91)/2]/-1 > {* (-1)} > sqrt(91)/2. Would this be correct?

Thanks,
Dan.

For reasons that mrspi explained to me but I have forgotten, it is traditional to eliminate surds from the denominators of fractions by mutiplying by a fraction that equals 1.

\(\displaystyle \dfrac{\sqrt{13}}{2\sqrt{7}} = \dfrac{\sqrt{13}}{2\sqrt{7}} * 1 = \dfrac{\sqrt{13}}{2\sqrt{7}} * \dfrac{\sqrt{7}}{\sqrt{7}} = \dfrac{\sqrt{13 * 7}}{2\sqrt{7 * 7}} = \dfrac{\sqrt{91}}{2 * 7} = \dfrac{\sqrt{91}}{14}.\)

 

[DanthePolishman]

\(\displaystyle \dfrac{a}{{1 + \sqrt b }} = \dfrac{{a(1 - \sqrt b) }}{{1 - b}}\) & \(\displaystyle \dfrac{c}{{\sqrt d }} = \dfrac{{c\sqrt d }}{d}\)

Let what be what?

 

[DanthePolishman]

For reasons that mrspi explained to me but I have forgotten, it is traditional to eliminate surds from the denominators of fractions by mutiplying by a fraction that equals 1.

\(\displaystyle \dfrac{\sqrt{13}}{2\sqrt{7}} = \dfrac{\sqrt{13}}{2\sqrt{7}} * 1 = \dfrac{\sqrt{13}}{2\sqrt{7}} * \dfrac{\sqrt{7}}{\sqrt{7}} = \dfrac{\sqrt{13 * 7}}{2\sqrt{7 * 7}} = \dfrac{\sqrt{91}}{2 * 7} = \dfrac{\sqrt{91}}{14}.\)

Thanks for the help. I think that my method could have gone right, but I have made an arithmetic problem somewhere, since I ended up with sqrt(91)/2.

 

[Quaid]

I think that my method could have gone right

I don't think so. Your method changes sqrt(7) to sqrt(-7).

When you did that, were you thinking that was using a conjugation?

sqrt(-7) is not a Real number.

Additionally, you later changed sqrt(-91) to sqrt(91) by multiplication with -1. That's also a mistake.

-1 times sqrt(number) = -sqrt(number)


Here is something else to consider, when asking for help on these boards.

You typed: 2/1 - sqrt(3)

This typing means \(\displaystyle 2-\sqrt{3}\) because we do division before subtraction.

To type what you actually meant, you must use grouping symbols to change the order of operations:

2/[1 - sqrt(3)]


If you're still wondering about anything in this thread, keep posting! Cheers :cool:

 

[DanthePolishman]

Thanks for all the help,

And I do have another question non-related to this thread. How do people notate math with real symbols? I.E. instead of writing sqrt(26) you actually have the number inside and actual square root picture.
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