Be p(x) = x³ + ax² + bx a polynomial with non negative roots and who are in an arithmetic progression. Knowing that the sum of the coefficients is 10, we can state that the sum of the roots of p(x) is?
Beautiful bird
what the fuck do i look like? your math teacher?
I'm the one asking questions here
but are you?
Yes, I'm the one asking questions here
Solution?
THEY DON'T ANSWER MY QUESTIONSmath stack exchange
you have to post and make it look like you tried then they'll answer
Nigga I think you're on the wrong course.
Wdym?
Asking us incels for solutions is surely no solution.
But I got many solutions here, I'm just waiting for a high IQcel to pop up and solve all problems for me in the PM, as it used to be
I got roots 0 an 0.488 giving a total sum of p(x) being 0.488
I found 3 roots was impossible and 1 root is not allowed
p(x)=x^3-8.325x^2+(-8.325)^2/4 x^3 (Approximate)
Wrong, it's 9
How is that possible?
One of the roots is zero.
a + b = 9.
Therefore: x² + ax + b.
Roots are in AP, therefore we can write it as:
A1, A1 + r, A1 + 2r.
A1 = 0, therefore:
0, 0 + r, 0 + 2r.
3r = -a.
b = 2r².
-3r + 2r² = 9.
2r² - 3r + -9 = 0.
r = 3 or -3/2.
But since the roots can't be negative, the value of r is 3.
Therefore 0, 3, 6.
So the answer is 9.
yh I dont know what the fuck i was doing
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