the angles are...90 degrees, 9-x^2, and 3x+38. is x possible? i've gotten to where -x^2+3x-43=0, but i can't do quadratic formula? as in it doesn't work? thanks!
2007-04-17 15:36:55 · 5 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
ok basically all the angles in a triangle are equal to 180o
so you have,
180 = 90 + (9-x^2) + (3x+38)
simplify
180 = 90 + 9 + 38 + 3x - x^2
rearrange
180= - x^2 + 3x + 137
arrange to equal zero
0 = -x^2 + 3x - 43
you should recongnise this as a quadratic equation
solve this and subsititute your answer's in to the original equation and you'll soon have your answer
2007-04-17 15:48:13 · answer #1 · answered by ~*~ 3 · 0⤊ 0⤋
Compute the discriminant of your quadratic:
b²-4ac = 9-4(43) < 0.
Since this is negative, there are no real roots,
so no such triangle is possible.
2007-04-17 22:49:12 · answer #2 · answered by steiner1745 7 · 0⤊ 0⤋
Looks impossible to me, since 9-(4)(43)<0. I would say there is no such triangle.
2007-04-17 22:44:25 · answer #3 · answered by airtime 3 · 1⤊ 0⤋
Yeh, I think you're right - there are no real roots for your formula. So, the answer must be no.
2007-04-17 22:41:34 · answer #4 · answered by Anonymous · 1⤊ 0⤋
No, it doesn't work.
There are no real solutions to that equation.. so such triangle exists.
2007-04-17 22:43:55 · answer #5 · answered by suesysgoddess 6 · 2⤊ 0⤋