Finding the Distance Between Two Points

[NoGoodAtMath]

I know the distance formula; d=√(x2-x1)^2 +(y2-y1)^2
So here is the problem:
(√5, -√2) and (4√5, -7√2)
d=√(4√5- √5)^2 + (-7√2- √2)^2

After this I have no idea. The square roots are tripping me up. What I did was took 3√5-5 and that became 15√5, then I multiplied 15(5) and got 75. Then I worked the other half. I got -8√2 -2 and that gave me and -32. I know I did something wrong here. Please me out. Thanks so much.

 

[HallsofIvy]

You don't see to be quite clear on how to work with radicals. First, -7√2- (-√2) is -7√2+ √2, NOT -7√2- √2.

But 4√5- √5 is certainly NOT "3√5- 5". And neither has anything to do with "15√5".

And -7√2- (-√2)= -7√2+ √2 is NOT "-8√2- 2".

You should see that this is really just integer arithmetic. What is 4x- x? What is -7x+ x?

 

[NoGoodAtMath]

You don't see to be quite clear on how to work with radicals. First, -7√2- (-√2) is -7√2+ √2, NOT -7√2- √2.

But 4√5- √5 is certainly NOT "3√5- 5". And neither has anything to do with "15√5".

And -7√2- (-√2)= -7√2+ √2 is NOT "-8√2- 2".

You should see that this is really just integer arithmetic. What is 4x- x? What is -7x+ x?

3x and -6x.
But the answer in the back of the book has √117.

 

[HallsofIvy]

3x and -6x.
But the answer in the back of the book has √117.

So now, what is "4√5- √5" and "-7√2+ √2"? What do you get when you put those into the formula?
Do the arithmetic one step at a time-
What is 4√5- √5? What is -7√2+ √2?
What is (4√5- √5)^2? What is (-7√2+ √2)^2?
What is (5√5- √5)^2+ (-7√2+ √2)^2?

Finally, what is √[(5√5- √5)^2+ (-7√2+ √2)^2]?

 

[pka]

So here is the problem:
(√5, -√2) and (4√5, -7√2)
d=√(4√5- √5)^2 + (-7√2- √2)^2

3x and -6x.
But the answer in the back of the book has √117.

\(\displaystyle \left( {\sqrt 5 ,-\sqrt 2 } \right)~\&~\left( {4\sqrt 5 , - 7\sqrt 2 } \right)\)

\(\displaystyle \sqrt {{{\left[ {\sqrt 5 - 4\sqrt 5 } \right]}^2} + {{\left[ { - \sqrt 2 + 7\sqrt 2 } \right]}^2}} = \sqrt {{{\left[ { - 3\sqrt 5 } \right]}^2} + {{\left[ {6\sqrt 2 } \right]}^2}} = \sqrt {9 \cdot 5 + 36 \cdot 2} \)

 

[NoGoodAtMath]

\(\displaystyle \left( {\sqrt 5 ,-\sqrt 2 } \right)~\&~\left( {4\sqrt 5 , - 7\sqrt 2 } \right)\)

\(\displaystyle \sqrt {{{\left[ {\sqrt 5 - 4\sqrt 5 } \right]}^2} + {{\left[ { - \sqrt 2 + 7\sqrt 2 } \right]}^2}} = \sqrt {{{\left[ { - 3\sqrt 5 } \right]}^2} + {{\left[ {6\sqrt 2 } \right]}^2}} = \sqrt {9 \cdot 5 + 36 \cdot 2} \)

I see what you did. You square the 3 and it turned into 9. The square root of √5 is 5! And the same goes from the 6 turns into 36 and √2 is just 2. I get it now! For some reason seeing it laid out like this made things a lot less confusing. I really appreciate you taking the time to answer my question. I'm really grateful. Thank you again!