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Mathematics 15 Online
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

jimthompson5910 (jim_thompson5910):

your set up is close, but it's incorrect unfortunately

jimthompson5910 (jim_thompson5910):

it should be this x^2+(2x+4)^2 = (2x+6)^2 notice how the "(2x+4)" is squared

OpenStudy (queenv):

How about is the next step correct? x^2+4x^2+16x+16=4x^2+24x+36

jimthompson5910 (jim_thompson5910):

yes it is, keep going

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

jimthompson5910 (jim_thompson5910):

close

jimthompson5910 (jim_thompson5910):

here's what I'm getting

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

OpenStudy (queenv):

How did you get the 20

jimthompson5910 (jim_thompson5910):

16 - 36 = -20

jimthompson5910 (jim_thompson5910):

on the left side, you had 16, but it somehow changed to 36

OpenStudy (queenv):

Did I skip steps/

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?

jimthompson5910 (jim_thompson5910):

-20 and -8, but yes

OpenStudy (queenv):

5*4 =20, 5+4=9, 2*10=20, 2+10 =12, 10+10=20 I have no idea

jimthompson5910 (jim_thompson5910):

how about -10 and 2

jimthompson5910 (jim_thompson5910):

-10 + 2 = -8 -10 * 2 = -20

OpenStudy (queenv):

ok, wasn't thinking of the negatives. (x-10)(x+8)?

jimthompson5910 (jim_thompson5910):

good

jimthompson5910 (jim_thompson5910):

oh sry misread

jimthompson5910 (jim_thompson5910):

it should be (x-10)(x+2)

OpenStudy (queenv):

(x-10)(x+2) that is what I meant sorry

jimthompson5910 (jim_thompson5910):

so if x^2-8x-20 = 0 then (x-10)(x+2) = 0

jimthompson5910 (jim_thompson5910):

and if (x-10)(x+2) = 0 then by the zero product property x-10 = 0 or x+2 = 0

OpenStudy (queenv):

that is far as we can go?

jimthompson5910 (jim_thompson5910):

no we can go a few steps further

jimthompson5910 (jim_thompson5910):

we need to fully isolate x

OpenStudy (queenv):

Ok, we haven't got that far yet, but will you show me?

jimthompson5910 (jim_thompson5910):

if x-10 = 0, then x = ???

OpenStudy (queenv):

0

jimthompson5910 (jim_thompson5910):

no

jimthompson5910 (jim_thompson5910):

if x-10 = 0, then x = ???

OpenStudy (queenv):

I0

jimthompson5910 (jim_thompson5910):

better

jimthompson5910 (jim_thompson5910):

if x+2 = 0, then x = ???

OpenStudy (queenv):

-2

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

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

jimthompson5910 (jim_thompson5910):

so that means we must toss x = -2 therefore, the only solution is x = 10

OpenStudy (queenv):

okay, thank you very much for your patience. May the Lord God Bless you, you have helped me greatly.

jimthompson5910 (jim_thompson5910):

you're welcome, glad to be of help

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