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Find X
2007-09-08 06:10:39 · 6 answers · asked by Kevin P 1 in Science & Mathematics ➔ Mathematics
Use the Pythagorean formula:
13² = (x - 5)² + (x + 2)²
Expand, combine terms, and solve the quadratic for x:
169 = x² - 10x + 25 + x² + 4x + 4
169 = 2x² - 6x + 29
2x² - 6x - 140 = 0
x² - 3x - 70 = 0
(x + 7)(x - 10) = 0
x = -7 or 10
Since a length cannot be negative, it follows that x must be greater than 5. Since -7 does not fulfill this requirement, we discard it and are left with the answer x = 10.
2007-09-08 06:22:21 · answer #1 · answered by computerguy103 6 · 0⤊ 0⤋
x=10
2007-09-08 13:14:06 · answer #2 · answered by Anonymous · 0⤊ 0⤋
13 ² = ( x - 5 ) ² + ( x + 2 ) ²
169 = x ² - 10 x + 25 + x ² + 4 x + 4
2 x ² - 6 x - 140 = 0
x ² - 3 x - 70 = 0
(x - 10) (x + 7) = 0
x = 10 is acceptable answer.
2007-09-12 04:11:12 · answer #3 · answered by Como 7 · 0⤊ 0⤋
By Pythagoras theorem, we get
(x - 5)^2 + (x + 2)^2 = 13^2
x^2 -10x + 25 + x^2 + 4x + 4 = 169
2x^2 - 6x + 29 = 169
2x^2 - 6x - 140 = 0
x^2 - 3x - 70 = 0
We a quadratic equation to solve,
x' = [3 + sqrt.(9 -4(-70))]/2 = [3 + sqrt.(289)]/2 = (3 + 17)/2
x' = 10
x'' = [3 - sqrt.(9 -4(-70))]/2 = [3 - sqrt.(289)]/2 = (3 - 17)/2
x'' = -7 (a negative value). We cannot have that.
So, the only solution is x = 10.
2007-09-08 13:25:54 · answer #4 · answered by Christine P 5 · 0⤊ 0⤋
(x+2)^2+(x-5)^2=13^2 (Pythogorean theorem)
x^2+4x+4+x^2-10x+25=169
2x^2-6x-140=0
x=10 or -7 (solving)
10 is the value as side cannot be negative
2007-09-08 13:53:13 · answer #5 · answered by cidyah 7 · 0⤊ 0⤋
X = 10 or -7
It cannot be negative so X = 10
2007-09-08 13:23:53 · answer #6 · answered by Nick H 2 · 0⤊ 0⤋