square roots

**thereddevils** · June 8th 2009, 04:17 AM

Without evaluating the square root or using the calculator , show that $\sqrt{18}+\sqrt{3}$ is less than 6 .

Thanks ,,

**masters** · June 8th 2009, 04:29 AM

Originally Posted by thereddevils

Without evaluating the square root or using the calculator , show that $\sqrt{18}+\sqrt{3}$ is less than 6 .

Thanks ,,

Hi thereddevils,

$\sqrt{18}+\sqrt{3} < 6$

Subtract $\sqrt{3}$ from both sides.

$\sqrt{18}<6-\sqrt{3}$

Square both sides.

$18<36-12\sqrt{3}+3$

$-21<-12\sqrt{3}$

Square both sides again.

$441>432$

**furor celtica** · June 8th 2009, 05:02 AM

ummmmm
isnt that wrong?

**CaptainBlack** · June 8th 2009, 05:19 AM

Originally Posted by thereddevils

Without evaluating the square root or using the calculator , show that $\sqrt{18}+\sqrt{3}$ is less than 6 .

Thanks ,,

$(\sqrt{18}+\sqrt{3})^2=21+2\sqrt{54}$

but:

$6^2=36=21+2\times 7.5=21+2 \sqrt{56.25}>21+2\sqrt{54}$

since $\sqrt{56.25}>\sqrt{54}$

Hence $6>\sqrt{18}+\sqrt{3}$

CB

**Soroban** · June 8th 2009, 05:21 AM

Hello, masters!

Sorry . . . you made a classic error.
(I'm familiar with it . . . I've made it enough times.)

. . $\sqrt{18}+\sqrt{3} \:<\: 6$

Subtract $\sqrt{3}$ from both sides:

. . $\sqrt{18}\:<\:6-\sqrt{3}$

Square both sides:

. . $18\:<\:36-12\sqrt{3}+3$

. . $-21\:<\:-12\sqrt{3}$

Square both sides again. . ← Here!

. . $441\:<\:432$ . . . . This is not true.

The error is perhaps the most insidious one.

When you squared, you multiplied the left side by $-21$ ,
. . the right by $-12\sqrt{3}$

And you know what happens . . .

**masters** · June 8th 2009, 05:25 AM

Originally Posted by Soroban

Hello, masters!

Sorry . . . you made a classic error.
(I'm familiar with it . . . I've made it enough times.)

The error is perhaps the most insidious one.

When you squared, you multiplied the left side by $-21$ ,
. . the right by $-12\sqrt{3}$

And you know what happens . . .

Lost my head. Still on my first cup of coffee this morning. Thanks.

**Math's-only-a-game** · June 8th 2009, 11:35 AM

Originally Posted by thereddevils

Without evaluating the square root or using the calculator , show that $\sqrt{18}+\sqrt{3}$ is less than 6 .

Thanks ,,

$ \left( {\sqrt {18} + \sqrt 3 < 6} \right) = \left( {\sqrt {2 \times 9} + \sqrt 3 < 6} \right) = $

$ = \left( {3\sqrt 2 + \sqrt 3 < 6} \right) = \left( {3(\sqrt 2 + \frac{{\sqrt 3 }} {3}) < 6} \right) = $

$ = \left( {(\sqrt 2 + \frac{{\sqrt 3 }} {3}) < 2} \right) = \left( {(\sqrt 2 + \frac{1} {{\sqrt 3 }}) < 2} \right) = $

$ = \left( {\frac{1} {{\sqrt 3 }} < 2 - \sqrt 2 } \right) = \left( {\frac{1} {{\sqrt 3 }} < 2(1 - \frac{{\sqrt 2 }} {2})} \right) $

Dividing by 2 and squaring both sides:

$ = \left( {\frac{1} {{2\sqrt 3 }} < (1 - \frac{1} {{\sqrt 2 }})} \right) = \left( {\frac{1} {{12}} < (\frac{3} {2} - \sqrt 2 )} \right) = $

$ - \frac{{17}} {{12}} < - \sqrt 2 ,\text{ - }\frac{{289}} {{144}} < - 2,\text{ }\frac{{289}} {{144}} > 2\text{ } $

**CaptainBlack** · June 8th 2009, 12:05 PM

Originally Posted by Math's-only-a-game

$ \left( {\sqrt {18} + \sqrt 3 < 6} \right) = \left( {\sqrt {2 \times 9} + \sqrt 3 < 6} \right) = $

$ = \left( {3\sqrt 2 + \sqrt 3 < 6} \right) = \left( {3(\sqrt 2 + \frac{{\sqrt 3 }} {3}) < 6} \right) = $

$ = \left( {(\sqrt 2 + \frac{{\sqrt 3 }} {3}) < 2} \right) = \left( {(\sqrt 2 + \frac{1} {{\sqrt 3 }}) < 2} \right) = $

$ = \left( {\frac{1} {{\sqrt 3 }} < 2 - \sqrt 2 } \right) = \left( {\frac{1} {{\sqrt 3 }} < 2(1 - \frac{{\sqrt 2 }} {2})} \right) $

$ = \left( {\frac{1} {{2\sqrt 3 }} < (1 - \frac{1} {{\sqrt 2 }})} \right) = \left( {\frac{1} {{12}} < (\frac{3} {2} - \sqrt 2 )} \right) = $

$ - \frac{{17}} {{12}} < - \sqrt 2 ,\text{ - }\frac{{289}} {{144}} < - 2,\text{ }\frac{{289}} {{144}} > 2\text{ } $

You need to use words to explain the steps in this (at least one is redundant).

CB

**Math's-only-a-game** · June 8th 2009, 12:17 PM

Originally Posted by CaptainBlack

You need to use words to explain the steps in this (at least one is redundant).

CB

I explained the step before the last, which is not that obvious imo. Anything else needing explaining please let me know. I omitted the explanation of steps on purpose as I find that we learn better when we struggle to work out things ourselves. However, your solution is the best of all and that's the one that should be taken into account, again imho.

**CaptainBlack** · June 8th 2009, 12:39 PM

Originally Posted by Math's-only-a-game

I explained the step before the last, which is not that obvious imo. Anything else needing explaining please let me know. I omitted the explanation of steps on purpose as I find that we learn better when we struggle to work out things ourselves. However, your solution is the best of all and that's the one that should be taken into account, again imho.

The last two steps need explanation, if we ignore the fact that at no point to you explain what you are attempting to do.

You are mis-using equality signs, if you must do it this way you should be using bi-implication " $\Leftrightarrow$ "

If you handed this in to me for marking I don't think you would get more than 10% of the available marks for it as it is.

CB

**VonNemo19** · June 8th 2009, 12:55 PM

Originally Posted by thereddevils

Without evaluating the square root or using the calculator , show that $\sqrt{18}+\sqrt{3}$ is less than 6 .

Thanks ,,

You could use approximation!

4.2426^2>18 and 1.75^2>3

4.2426+1.75<6

there fore $\sqrt{18}+\sqrt{3}<6$

p.s. I assume that you know how to find approximate roots by averaging. I had a whole presentation on maple, but MHF will not let me upload. Maybe this'll help

p.s.s I only had to repeat twice for the approximation of root18 and once for root3. It only took a second!

p.s.s.s. If you would like to see my maple presentation of this concept, send me a pm and I'll send it to you.
[edit] Babylonian method

Graph charting the use of the Babylonian method for approximating the square root of 100 (10) using start values x0=50, x0=1, and x0=-5. Note that using a negative start value yields the negative root.

Perhaps the first algorithm used for approximating

is known as the "Babylonian method", named after the Babylonians,[1] or "Heron's method", named after the first-century Greek mathematician Heron of Alexandria who gave the first explicit description of the method.[2] It can be derived from (but predates) Newton's method. This is a quadratically convergent algorithm, which means that the number of correct digits of the approximation roughly doubles with each iteration. It proceeds as follows:

Start with an arbitrary positive start value x0 (the closer to the root, the better).
Let xn+1 be the average of xn and S / xn (using the arithmetic mean to approximate the geometric mean).
Repeat steps 2 and 3, until the desired accuracy is achieved.

It can also be represented as:

This algorithm works equally well in the p-adic numbers, but cannot be used to identify real square roots with p-adic square roots; it is easy, for example, to construct a sequence of rational numbers by this method which converges to + 3 in the reals, but to − 3 in the 2-adics.

**furor celtica** · June 8th 2009, 09:16 PM

where was masters mistake in squaring both sides?

**mathaddict** · June 8th 2009, 09:21 PM

Originally Posted by furor celtica

where was masters mistake in squaring both sides?

Read Soroban's post . It affects the inequality sign .

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