how do u solve:
x^2 +41x=40
?????? please help!
2006-11-29 13:07:32 · 5 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
x^2+41x-40=0
use quadratic formula.
2006-11-29 13:09:41 · answer #1 · answered by Poetic Disaster 1 · 0⤊ 0⤋
x^2 + 41x - 40 = 0
First, figure out what 40 multiples out from. For example, 9 would be 3 times 3. So we now those numbers in reference to the 41x because we are trying to find two numbers that add together to form 41, yet multiple together to form 40. unfortunately this is not a nice question so you must now follow the formula x = b +- Square Root of ( b^2 - 4ac) end Square Root, divided by 2a.
Where aX^2 + bX + c = 0
2006-11-29 13:13:35 · answer #2 · answered by nobleisback 1 · 0⤊ 0⤋
First subtract 41 from both sides of the problem equals: x^2+x=(-1) then add x^2 and x to get: x^3=(-1) next since the answer is (-1) put (-1) for x's equivilent x=(-1)
Hope this helped
2006-11-29 13:25:14 · answer #3 · answered by steamhead10 2 · 0⤊ 0⤋
x^2 + 41x = 40
x^2 + 41x - 40 = 0
x = (-b ± sqrt(b^2 - 4ac))/(2a))
x = (-41 ± sqrt(41^2 - 4(1)(-40)))/(2(1))
x = (-41 ± sqrt(1681 + 160))/2
x = (-41 ± sqrt(1841))/2
x = (1/2)(-41 ± sqrt(1841))
Assuming you didn't type anything wrong, that was what i got.
2006-11-29 13:15:58 · answer #4 · answered by Sherman81 6 · 0⤊ 0⤋
are you sure you have the signs right?????
2006-11-29 13:12:31 · answer #5 · answered by Tony T 4 · 0⤊ 0⤋