please help

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czagara

Junior Member
Joined
Sep 24, 2009
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Find x such that the distance between the points is 15.
(x, 11), (–6, 23)

please tell me what i am doing wrong here.

all of this is under a square root symbol:

(x+6)^2 (11-23)^2 =15

(x+6)^2 +144=225

subtract 144 from both = sides

(x+6)^2 = 81

and then i am lost.....
 
Everything you have done is correct and you are here:

\(\displaystyle (x+6)^2 = 81\)

Take the square root of both sides:

\(\displaystyle \sqrt{(x+6)^2} = \sqrt{81}\)
\(\displaystyle x+6 = \pm9\)
\(\displaystyle x = 3, -15\)

Hope this helps!

Also, please note that it might be easier to get help if you make a more descriptive title, unlike "Please help".
 
3 helped but its asking for a smaller value than 3
like it wants 2 answers, one is bigger than the other
 
czagara said:
Find x such that the distance between the points is 15.
(x, 11), (–6, 23)

please tell me what i am doing wrong here.

all of this is under a square root symbol:

(x+6)^2 + (11-23)^2 =15[sup:14ytdwkj]2[/sup:14ytdwkj]

(x+6)^2 +144=225

subtract 144 from both = sides

(x+6)^2 = 81

and then i am lost.....
Click to expand...
\(\displaystyle (x + 6)^2 = 81\)

\(\displaystyle [(x + 6)^2]^{\frac{1}{2}} = [81]^{\frac{1}{2}}\)

\(\displaystyle x + 6 = \pm 9\)
 
Juicy Burger forgot that the square root of (x + 6)^2 is not x + 6.

It's the absolute value of x + 6.

|x + 6| = 9

To remove the absolute value symbols, we need to consider both roots.

x + 6 = 9

or

x + 6 = -9
 
i got ya. thanks that makes sense and thankfully i got that one right.
 
Another way:

(x+6)[sup:lkg1edzq]2[/sup:lkg1edzq] - 81 = 0

x[sup:lkg1edzq]2[/sup:lkg1edzq] + 12x - 45 = 0

(x + 15)(x - 3) = 0 ? x = -15 or x = 3
 
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