# Equation needs URGENT checking

• Feb 21st 2010, 02:21 AM
99.95
Equation needs URGENT checking
Hello, i was preparing for a test i have tomorrow so was doing review questions in my text book. I came across this one question where i knew what to do but got the wrong answer, must be a mistake in my working out
Question: the distance between (2,-7) and (a,-2) is $\displaystyle \sqrt{41}$units. Find the value of a if it is positive.

Edit: New to using Latex - sorry for any inconvenience
AND
The answer i have been obtaining is a=4

The formula to use was d = $\displaystyle \sqrt(x2-x1)^2+(y2-y1)^2$
in this case d being $\displaystyle \sqrt 41$
the actual answer is a= 6..

$\displaystyle \sqrt41 = \sqrt{(a-2)^2+(-2+7)^2} \Rightarrow \sqrt41 = \sqrt{(a^2-4a+29)} \Rightarrow 41 = a^2-4a+29$
$\displaystyle \Rightarrow 12 = a^2-4a \Rightarrow 12/a = a-4$
$\displaystyle \Rightarrow 12/a+4 = a \Rightarrow 12+4 = a^2$
$\displaystyle \Rightarrow 16= a^2 \Rightarrow a=4$
• Feb 21st 2010, 02:29 AM
HallsofIvy
Quote:

Originally Posted by 99.95
Hello, i was preparing for a test i have tomorrow so was doing review questions in my text book. I came across this one question where i knew what to do but got the wrong answer, must be a mistake in my working out
Question: the distance between (2,-7) and (a,-2) is [tex]\sqrt{41}[\math]units. Find the value of a if it is positive.

Edit: New to using Latex, why wont it show upo

because you end with /math, not \math.

Quote:

AND
The answer i have been obtaining is a=4

The formula to use was d = \sqrt(x2-x1)^2+(y2-y1)^2
in this case d being \sqrt 41
the actual answer is a= 6..
What do you want? If you know the answer is 6, then you know your answer is wrong. If you want us to check your work and see where you went wrong, you will need to show your work!
• Feb 21st 2010, 02:33 AM
99.95
Quote:

Originally Posted by HallsofIvy
because you end with /math, not \math.

What do you want? If you know the answer is 6, then you know your answer is wrong. If you want us to check your work and see where you went wrong, you will need to show your work!

ok ill post it now, yea i want you to check and see where i went wrong :)

$\displaystyle \sqrt41 = \sqrt{(a-2)^2+(-2+7)^2} \Rightarrow \sqrt41 = \sqrt{(a^2-4a+29)} \Rightarrow 41 = a^2-4a+29 \Rightarrow 12 = a^2-4a \Rightarrow 12/a = a-4 \Rightarrow 12/a+4 = a \Rightarrow 12+4 = a^2 \Rightarrow 16= a^2 \Rightarrow a=4$
• Feb 21st 2010, 02:40 AM
99.95
$\displaystyle \sqrt41 = \sqrt{(a-2)^2+(-2+7)^2} \Rightarrow \sqrt41 = \sqrt{(a^2-4a+29)} \Rightarrow 41 = a^2-4a+29$
$\displaystyle \Rightarrow 12 = a^2-4a \Rightarrow 12/a = a-4$
$\displaystyle \Rightarrow 12/a+4 = a \Rightarrow 12+4 = a^2$
$\displaystyle \Rightarrow 16= a^2 \Rightarrow a=4$
• Feb 21st 2010, 04:06 AM
Wilmer
WHAT are you doing?!
You have a^2 - 4a + 29 = 41; so:
a^2 - 4a - 12 = 0
(a - 6)(a + 2) = 0
It's that simple (Cool)
• Feb 21st 2010, 04:25 AM
99.95
Quote:

Originally Posted by Wilmer
WHAT are you doing?!
You have a^2 - 4a + 29 = 41; so:
a^2 - 4a - 12 = 0
(a - 6)(a + 2) = 0
It's that simple (Cool)

Wow, even though it was so simple, i would have never gotten that in a million years..
Too rusty for my own good.
thanks though!
• Feb 21st 2010, 04:54 AM
Wilmer
Never too late; pick a couple sites here...become an expert!