Finding variables using the distance equation

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Solar_blaze

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I almost resolved this question need to check if I got it right
"Find a coordinate point on the axis of X , which it's distance from the point(8:7) is 25.
I did this
25=(8-X)^2+(7-0)^2 than
25=64-16X+X^2+49 than
25^2=113-16X+X^2
625=113-16X+X^2 than
113-625-16X+X^2=0
-512-16X+X^2
If I understand it right now I need to place it inside the " quadratic equation"?
 
Solar_blaze said:
I almost resolved this question need to check if I got it right
"Find a coordinate point on the axis of X , which it's distance from the point(8:7) is 25.
I did this
25=(8-X)^2+(7-0)^2 than
25=64-16X+X^2+49 than
25^2=113-16X+X^2
625=113-16X+X^2 than
113-625-16X+X^2=0
-512-16X+X^2
If I understand it right now I need to place it inside the " quadratic equation"?
Click to expand...

Yes..
 
Solar_blaze said:
I almost resolved this question need to check if I got it right
"Find a coordinate point on the axis of X , which it's distance from the point(8:7) is 25.
I did this
25=(8-X)^2+(7-0)^2 than
25=64-16X+X^2+49 than
25^2=113-16X+X^2
625=113-16X+X^2 than
113-625-16X+X^2=0
-512-16X+X^2
If I understand it right now I need to place it inside the " quadratic equation"?
Click to expand...

I'll quibble a bit with your notation.....
I think your first two lines should look like this:

25 = sqrt[(8 - x)2 + (7 - 0)2]
25 = sqrt[64 - 16x + x2 + 49]
THEN....square both sides to get your third line:
252 = 113 - 16x + x2

Then you are good to go, and yes you would want to use the quadratic formula (or factor....this isn't going to be too difficult to factor).
 
Solar_blaze said:
When you square 25 it's 625 NoT 252
Click to expand...

He meant 25^2.

Your answers should be x= -16, x = 32.

It's really important to learn to factor without the quadratic equation, you should dig into that subject. The quadratic equation is only needed if there are only irrational or imaginary solutions.
 
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