Length of a Hypotenuse (symbolic algebra)

[Anthobet]

Using Pythagoras c² = a² + b² I substituted my values and obtained the following:

c² = (a² - b²)² + (2ab)²

After distributing and combining like terms, I obtain

c = √[a⁴ + 2a²b² + b⁴]

I couldn't figure out how to algebraically manipulate this to match any of the answers in the photo above. So I selected something that closely resembles my answer. Thank you for reading.​

 

[Dr.Peterson]

Try squaring each of the choices to see if it matches your radicand. Then think about how you might have recognized that on your own. (It's a good thing to discover.)

Resemblance is quite different from equality.

 

[Otis]

Hello Anthobet. I'm thinking that Dr. Peterson may have one of the special factoring patterns in mind:

(x + y)^2 = x^2 + 2xy + y^2

In words, (1) we square each term, (2) we multiply the two terms and double that product, and then (3) we combine the three results to finish.

The binomial above is x + y, so the terms are x and y.

(1) Square each term: x^2 and y^2

(2) Multiply the terms and double: 2xy

(3) Combine the results: x^2 + 2xy + y^2

Of course, one could square (x+y) by applying the FOIL algorithm (aka: double distribution) -- and maybe you should (to see for yourself) -- but remembering the special factoring pattern is often easier.

(3a - 7b)^2

Square each term (9a^2 and 49b^2), multiply the terms and double (-21ab) and combine the results:

9a^2 - 21ab + 49b^2 = (3a - 7b)^2

With some practice, you can apply this special factoring pattern in your head.

And, once you're comfortable with the pattern, you can use it in reverse. That is, if you have a trinomial of form x^2 + 2xy + y^2, then you know right away that it factors as the square of a binomial: (x+y)^2.

Dr. Peterson suggested you think about the form x^4 + 2*x^2*y^2 + y^4. Does that give you any ideas?

?

 

[Anthobet]

Try squaring each of the choices to see if it matches your radicand. Then think about how you might have recognized that on your own. (It's a good thing to discover.)

Resemblance is quite different from equality.

I CAN'T BELIEVE MY EYES! That is so cool! For the longest time, I've remembered (a + b)² = a² + 2ab + b² and (a - b)² = a² - 2ab + b² and I can't believe this identity went past my eyes!

When you asked me to square each of the choices I thought I would discover some new type of math that can only be unraveled from working out the given answer. And it blows my mind how the problem circled back to Pythagoras.

So final solution:
c² = a⁴ + 2a²b² + b⁴ = (a² + b²)²
then
c = √[(a² + b²)²]
c = a² + b²

Thank you so much Dr.Peterson

[edit] Thank you too for replying as well Otis.

 

[Dr.Peterson]

Now, notice something interesting: adding the square of 2ab changes the sign in (a^2 - b^2)^2 from - to +, by doing the same to the middle term of the expansion. This sort of thing happens more than you'd think, so, as I said, it's a good thing to have seen.