Mathematics
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OpenStudy (queenv):
Buried treasure. Ahmed has half of a treasure map, which indicates that the treasure is buried in the desert 2x+6 paces from Castle Rock. Vanessa has the other half of the map. Her half indicates that to find the treasure, one must get to Castle Rock, walk x paces to the north, and then walk 2x+4 paces to the east. If they share their information, then they can find x and save a lot of digging. What is x?
Pythagorean Quadratic problem. Can some one please tell me if I set this problem up correct?
x^2+(2x+4) = (2x+6)^2
12 years ago
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jimthompson5910 (jim_thompson5910):
your set up is close, but it's incorrect unfortunately
12 years ago
jimthompson5910 (jim_thompson5910):
it should be this
x^2+(2x+4)^2 = (2x+6)^2
notice how the "(2x+4)" is squared
12 years ago
OpenStudy (queenv):
How about is the next step correct? x^2+4x^2+16x+16=4x^2+24x+36
12 years ago
jimthompson5910 (jim_thompson5910):
yes it is, keep going
12 years ago
OpenStudy (queenv):
-4x^2 -4x^2
x^2+16x+16=24x+36
-24x -24x
x^2-8x+36=36
-36 -36
x^2-8x-0+0
12 years ago
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jimthompson5910 (jim_thompson5910):
close
12 years ago
jimthompson5910 (jim_thompson5910):
here's what I'm getting
12 years ago
jimthompson5910 (jim_thompson5910):
x^2+(2x+4)^2 = (2x+6)^2
x^2+(2x+4)(2x+4) = (2x+6)(2x+6)
x^2+2x(2x+4)+4(2x+4) = 2x(2x+6)+6(2x+6)
x^2+4x^2+8x+8x+16 = 4x^2+12x+12x+36
5x^2+16x+16 = 4x^2+24x+36
5x^2+16x+16 - 4x^2-24x-36 = 0
x^2-8x-20 = 0
12 years ago
OpenStudy (queenv):
How did you get the 20
12 years ago
jimthompson5910 (jim_thompson5910):
16 - 36 = -20
12 years ago
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jimthompson5910 (jim_thompson5910):
on the left side, you had 16, but it somehow changed to 36
12 years ago
OpenStudy (queenv):
Did I skip steps/
12 years ago
OpenStudy (queenv):
Ok I got you. I see. x^2-8x-20 so now I need two factors of 20 that add up to 8?
12 years ago
jimthompson5910 (jim_thompson5910):
-20 and -8, but yes
12 years ago
OpenStudy (queenv):
5*4 =20, 5+4=9, 2*10=20, 2+10 =12, 10+10=20 I have no idea
12 years ago
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jimthompson5910 (jim_thompson5910):
how about -10 and 2
12 years ago
jimthompson5910 (jim_thompson5910):
-10 + 2 = -8
-10 * 2 = -20
12 years ago
OpenStudy (queenv):
ok, wasn't thinking of the negatives. (x-10)(x+8)?
12 years ago
jimthompson5910 (jim_thompson5910):
good
12 years ago
jimthompson5910 (jim_thompson5910):
oh sry misread
12 years ago
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jimthompson5910 (jim_thompson5910):
it should be (x-10)(x+2)
12 years ago
OpenStudy (queenv):
(x-10)(x+2) that is what I meant sorry
12 years ago
jimthompson5910 (jim_thompson5910):
so if
x^2-8x-20 = 0
then
(x-10)(x+2) = 0
12 years ago
jimthompson5910 (jim_thompson5910):
and if
(x-10)(x+2) = 0
then by the zero product property
x-10 = 0 or x+2 = 0
12 years ago
OpenStudy (queenv):
that is far as we can go?
12 years ago
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jimthompson5910 (jim_thompson5910):
no we can go a few steps further
12 years ago
jimthompson5910 (jim_thompson5910):
we need to fully isolate x
12 years ago
OpenStudy (queenv):
Ok, we haven't got that far yet, but will you show me?
12 years ago
jimthompson5910 (jim_thompson5910):
if x-10 = 0, then x = ???
12 years ago
OpenStudy (queenv):
0
12 years ago
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jimthompson5910 (jim_thompson5910):
no
12 years ago
jimthompson5910 (jim_thompson5910):
if x-10 = 0, then x = ???
12 years ago
OpenStudy (queenv):
I0
12 years ago
jimthompson5910 (jim_thompson5910):
better
12 years ago
jimthompson5910 (jim_thompson5910):
if x+2 = 0, then x = ???
12 years ago
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OpenStudy (queenv):
-2
12 years ago
jimthompson5910 (jim_thompson5910):
so this means we have these 4 steps at the end
x^2-8x-20 = 0
(x-10)(x+2) = 0
x-10 = 0 or x+2 = 0
x = 10 or x = -2
12 years ago
jimthompson5910 (jim_thompson5910):
but remember that x is a distance or length
you CANNOT have negative distances or lengths, so x = -2 makes no sense in the real world
12 years ago
jimthompson5910 (jim_thompson5910):
so that means we must toss x = -2
therefore, the only solution is x = 10
12 years ago
OpenStudy (queenv):
okay, thank you very much for your patience. May the Lord God Bless you, you have helped me greatly.
12 years ago
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jimthompson5910 (jim_thompson5910):
you're welcome, glad to be of help
12 years ago