Approximating the value of the square root of 101

[MichaelOB]

The question is:

"Show that [MATH]\sqrt{N+1} - \sqrt{N} = \frac{1}{\sqrt{N+1}+\sqrt{N}}[/MATH]. Use this to explain why [MATH]\sqrt{101}[/MATH] is close to but slightly less that, 10.05.
Without using a calculator, find the roots of [MATH]x^2 +7x - 13[/MATH], giving your answers correct to 2 d.p"

The picture shows my working. My first problem is that I don't think I have given a tight enough interval which shows that the value is close to 10.05.

Secondly, the second part of the question involves the square root of 101 and I round the roots up because the root are 'close to 10.05. However, the answer in the back of the book gives the answers 1.52 and -8.52. I would probably need a smaller interval for the value of the root of 101 and that leads back to my first problem.

What do I need to do to answer this question accurately. Thanks.

 

[Dr.Peterson]

I think you've done all you need to do on the first part. It's less than 10.05, and can't be much less because the denominator is not much more than 20. If you replace [MATH]\sqrt{101}[/MATH] with 10.05 in the formula, you get [MATH]\frac{1}{20.05}\approx 0.049875[/MATH], which is not much less than 0.05. If you were expected to do more, the wording would have made it clear (defining "close" and "slightly").

As for the quadratic, you just forgot the inequality. Since [MATH]\sqrt{101}< 0.05[/MATH], you need to round 1.525 down (as it's a little less than 1.525), and similarly you have to round -8.525 up (to -8.52).

 

[Steven G]

You did not round off correctly

 

[pka]

The question is: "Show that [MATH]\sqrt{N+1} - \sqrt{N} = \frac{1}{\sqrt{N+1}+\sqrt{N}}[/MATH]. Use this to explain why [MATH]\sqrt{101}[/MATH] is close to but slightly less that, 10.05. Without using a calculator, find the roots of [MATH]x^2 +7x - 13[/MATH], giving your answers correct to 2 d.p"

When ever I still see these sorts of question, I must wonder why someone still whats back to the basics.
At one point we did need this sort of approximation. But this is the twenty first century. We have computer algebra systems.
I can see why we may want calculus students to know that \(\displaystyle \sqrt {N + 1} - \sqrt N = \frac{1}{{\sqrt {N + 1} + \sqrt N }}\)
for the lessons on limits of indeterminate forms. OR for a course in the history of mathematics.
The Keith Devlin piece is twenty years old, lets move on.

 

[JeffM]

When ever I still see these sorts of question, I must wonder why someone still whats back to the basics.
At one point we did need this sort of approximation. But this is the twenty first century. We have computer algebra systems.
I can see why we may want calculus students to know that \(\displaystyle \sqrt {N + 1} - \sqrt N = \frac{1}{{\sqrt {N + 1} + \sqrt N }}\)
for the lessons on limits of indeterminate forms. OR for a course in the history of mathematics.
The Keith Devlin piece is twenty years old, lets move on.

First, thank you for the link.

Second, I am sympathetic to the argument but far from persuaded. No one needs to know arithmetic in the 21st century; calculators are cheap and portable. Yet I greatly doubt that whatever deeper mathematical ideas that Devlin wants to have taught instead of addition will be comprehensible to adolescents who cannot add five and three without a calculator. (I think we can agree that addition is basic unless we want to introduce the Peano postulates in first grade.)

So for example, I agree with you that the utility of logs to the base 10 to do computation has been eliminated over the last fifty years. Logs to the base 10, however, remain the easiest way to introduce the whole family of logarithmic functions to students. I'd never introduce an algebra student to logs through "natural" logs despite their central role in calculus. On the other hand, I'd also never teach a student today how to do interpolations in a log table.

 

[MichaelOB]

I think you've done all you need to do on the first part. It's less than 10.05, and can't be much less because the denominator is not much more than 20. If you replace [MATH]\sqrt{101}[/MATH] with 10.05 in the formula, you get [MATH]\frac{1}{20.05}\approx 0.049875[/MATH], which is not much less than 0.05. If you were expected to do more, the wording would have made it clear (defining "close" and "slightly").

As for the quadratic, you just forgot the inequality. Since [MATH]\sqrt{101}< 0.05[/MATH], you need to round 1.525 down (as it's a little less than 1.525), and similarly you have to round -8.525 up (to -8.52).

Thank you Dr. Peterson.