find length of 3rd side of triangle; explain error in simplification; 2/1+sqrt 2 = ?

[Dot_01]

I've been stuck on the following 3 problems for a while they are: 1) If a right triangle has a hypotenuse of 5 sqrt 5 cm and one side that is twice as long as the other side, what is the length of the shortest side? A) 3 sqrt 5 B)5 C) 25 and D) 15.

2) [FONT=&quot]Explain the error in this simplification of radical expressions. What is the correct simplification? sqrt 2 * 3sqrt8= sqrt 2(8)=sqrt 16=4. A)The product of two radical expressions can never be expressed as one single radical expression B)The Product Property does not apply to different indices. C)The product property does not apply to different radicans and D) When multiplying two radical expressions together, the index of the product should be the sum of the indices of the multiplicands.

3) 2/1+sqrt 2= A) 1/1- sqrt 2, B) 2-sqrt2/2, c) -2-2sqrt2, D) 2+2sqrt2. (For this one, I keep getting -2+2sqrt2, I wonder if i'm messing up anywhere.) [/FONT][FONT=&quot][FONT=&quot][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main]√[/FONT][FONT=MathJax_Main]∙[/FONT][FONT=MathJax_Main]8[/FONT][FONT=MathJax_Main]√[/FONT][FONT=MathJax_Main]3[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Main]8[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Size1]√[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]16[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]√[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]4[/FONT][/FONT][/FONT]

 

[stapel]

1) If a right triangle has a hypotenuse of 5 sqrt 5 cm and one side that is twice as long as the other side, what is the length of the shortest side? A) 3 sqrt 5 B)5 C) 25 and D) 15.

Draw a generic right triangle. Label the hypotenuse with the given value. The longer side is defined in terms of the shorter side, so pick a variable for the shorter side, and create an expression for the longer side. Label the drawing. Then apply the Pythagorean Theorem.

2) Explain the error in this simplification of radical expressions. What is the correct simplification? sqrt 2 * 3sqrt8= sqrt 2(8)=sqrt 16=4. A)The product of two radical expressions can never be expressed as one single radical expression B)The Product Property does not apply to different indices. C)The product property does not apply to different radicans and D) When multiplying two radical expressions together, the index of the product should be the sum of the indices of the multiplicands.

I'm not clear on what you've posted...? By "sqrt 2 * 3sqrt8= sqrt 2(8)=sqrt 16=4", did you mean the following?

. . .$\displaystyle \sqrt{2\,} \, \cdot\, 3\, \sqrt{8\,}$

. . . . .$\displaystyle =\, \sqrt{2(8)\,}\, =\, \sqrt{16\,}\, =\, 4$

If so, what happened to the "3" that went with the second radical?

3) 2/1+sqrt 2= A) 1/1- sqrt 2, B) 2-sqrt2/2, c) -2-2sqrt2, D) 2+2sqrt2. (For this one, I keep getting -2+2sqrt2, I wonder if i'm messing up anywhere.) 2√∙8√3=2(8)−−−−√=16−−√=4

What you have posted means the following:

. . . . .$\displaystyle \dfrac{2}{1}\, +\, \sqrt{2\,}$

I suspect you actually mean the following:

. . . . .$\displaystyle \dfrac{2}{1\, +\, \sqrt{2\,}}$

Please confirm or correct.

I've been stuck on the following 3 problems for a while

Please reply showing at least one of your efforts for each of the exercises, so we can see what's going on and where you're getting stuck. Thank you!

 

[Dr.Peterson]

2) Explain the error in this simplification of radical expressions. What is the correct simplification? sqrt 2 * 3sqrt8= sqrt 2(8)=sqrt 16=4. A)The product of two radical expressions can never be expressed as one single radical expression B)The Product Property does not apply to different indices. C)The product property does not apply to different radicans and D) When multiplying two radical expressions together, the index of the product should be the sum of the indices of the multiplicands.

[FONT=MathJax_Main]2[FONT=MathJax_Main]√[FONT=MathJax_Main]∙[/FONT][FONT=MathJax_Main]8[/FONT][FONT=MathJax_Main]√[/FONT][FONT=MathJax_Main]3[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Main]8[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Size1]√[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]16[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]√[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]4[/FONT][/FONT][/FONT]

Based on the choices, I suspect that when you wrote "3 sqrt8" you didn't mean what it says (3 times the square root of 8), but rather "the cube root of 8", $\displaystyle \sqrt[3]{8}$. (There is no fully standard way to write this in plain text; I would write "cbrt(8)".)

That last line I quoted above looks like your attempt to show the keystrokes you used on some particular calculator to evaluate the expression for (2); but I can't be sure from that whether you mean a cube root or 3 times a square root. In any case, the 3 seems not to have been used.

 

[JeffM]

(There is no fully standard way to write this in plain text; I would write "cbrt(8)".)

@ Dot

Minor point that does not disagree with anything material in Dr. Peterson's post. But it may be helpful to you to know that you can easily describe in standard form the nth root of x in plain text by x^(1/n) because

$\displaystyle \sqrt[n]{x} \equiv x^{1/n}.$

 

[Dot_01]

Draw a generic right triangle. Label the hypotenuse with the given value. The longer side is defined in terms of the shorter side, so pick a variable for the shorter side, and create an expression for the longer side. Label the drawing. Then apply the Pythagorean Theorem.


I'm not clear on what you've posted...? By "sqrt 2 * 3sqrt8= sqrt 2(8)=sqrt 16=4", did you mean the following?

. . .$\displaystyle \sqrt{2\,} \, \cdot\, 3\, \sqrt{8\,}$

. . . . .$\displaystyle =\, \sqrt{2(8)\,}\, =\, \sqrt{16\,}\, =\, 4$

If so, what happened to the "3" that went with the second radical?


What you have posted means the following:

. . . . .$\displaystyle \dfrac{2}{1}\, +\, \sqrt{2\,}$

I suspect you actually mean the following:

. . . . .$\displaystyle \dfrac{2}{1\, +\, \sqrt{2\,}}$

Please confirm or correct.


Please reply showing at least one of your efforts for each of the exercises, so we can see what's going on and where you're getting stuck. Thank you!

For the second one, no clue as to what happened to the index, but I figured out what the error was. For the 1st one about the triangle, I tried setting the pythagorean theorem to having the answer choice(s) as A, whatever A squared was as B to equal 5 sqrt 5 (which is 125.) So for example, this equation would be (5)^2+(25)^2=125. I did this for every available answer choice (A being 3 sqrt 5, B being 5, C being 25 and D being 15. After substituting each answer choice like this into the equation, I have found that none of them are true.

For the 3rd question, what you assumed it was is correct (2 divided by 1 plus the square root of 2.) The work I have is I multiply by the conjugate 1 minus the square root of 2 over 1 minus the square root of -2 to get 2(1-sqrt 2) over 1. I then apply the rule A/1=a and get -2(1-sqrt 2), distribute, and get -2+2sqrt2.