₦airaland Forum

FunnyDude:
Taaar! If I hear.

respect80:
Drop your whatsapp number if you need the mtn simple server configuration

naturalwaves:
Let me help you....

goofyone:
If -i is a root, i will also be a root following the complex conjugate theorem. I can prove that if you want, but i don't think it's relevant to do that here.

Hence, two roots already, ± i

(x+i)(x-i) = x² -i.i = x² + 1

x⁴ + 4x³ + 6x² + 4x + 5 divided by x² + 1 to get the other factor.

Division produces x² + 4x + 5

Solving with quadratic equation: ¹/₂(-4 ± √(16 - 20))
x = -2 ± i

goofyone:
| 1 1 1 |
| x y z |
| x² y² z² |

= (y.z² - z.y²) - (x.z² - z.x²) + (x.y² - y.x²)
=zy(z-y)-zx(z-x)+xy(y-x)
=z(y(z-y) - x(z-x)) + xy(y-x)
=z(yz-xz + x² - y²) + xy(y-x)
=z(z(y-x)+(x-y)(x+y))+xy(y-x)
=z(z(y-x)-(y-x)(x+y))+xy(y-x)
=z(y-x).(z-(x+y))+xy(y-x)
=(y-x)(z.(z-(x+y)+xy)
=(y-x)(z²-zx-zy+xy)
=(y-x)(z²-zy+xy-zx)
=(y-x)(z(z-y)-x(z-y))
=(y-x)(z-x)(z-y)

naturalwaves:
Check the first question well, there are 2 functions there, which of them do we go with? I guess it is the second one but quickly review it or take a snapshot of the question....waiting....

goofyone:
For this question, you can use Descartes's rule of signs. You should actually use the more detailed Sturm's extension of this rule, but i think the former could be okay here.
Descartes's rule of signs says if v is the variation of sign in a polynomial p(x) and the number of positive real zeros (roots) is n, then
n<=v and v-n is even.

Same applies to number of negative real zeros with p(-x)

With m=1, you get
x³ + x² + 5x + 3 = 0


p(x) = x³ + x² + 5x + 3
No sign variation, so v=0.
n will also be 0 so there are no positive real roots.

p(-x) = -x³ + x² - 5x + 3

Number of sign variation is 3 (- to + to - to +)
n will be 1, so that 3-1 is even. So there's a negative real root.

Equation thus has one negative real root and two complex roots which will be conjugates.


With m= -²/₃

p(x) = -(²/₃)x³ - (⁷/₃)x² + 3

Number of sign variation is 1. n will be 1 so that 1-1=0 which is even.
Hence, one positive real root.

p(-x) = (²/₃)x³ - (⁷/₃)x² + 3.

Number of sign variation here is 2. n cannot be 1, cos v-n should be even. So n is 2. And 2-2=0.
So we have two negative real roots.

Hence, three real roots (one positive and two negative)

frob0genius:
I do have it. How can I send it to you?
It's over 10mb

Midehi:
you curly hair na, thats what i want to see,upload ur profile pics jor

Midehi:
wish i can see the hair

deb:
Haters, he was declared winner on April 1st, 2015.

Meanwhile, what do you all ranters have to say now that GEJ's Army suddenly found Buhari's certificate after he was declared a winner? Na God go punish all of you wey no ask for forgiveness for aligning with the most corrupt Nigerian government ever.

"Ees our brother..." "but eees people that are sabotaging him" "Ees a good man" bla bla bla Eeddiaats!