Question

How many roots do the following equations have?
-12x2 - 25x + 5 + x3 = 0
A. 3
B. 4
C. 5
D. 2
it's not D.

Answer 1

@DatChinookGuy can you help

Answer 2

I'd say 3

Answer 3

how did you get 3

Answer 4

Basically, the number of roots for an equation is equal to the highest power in the equation.
So, quadratics have 2
cubic equations have 3
and so on

Answer 5

okay thanks that helped alot. :)

Answer 6

u r welcome

Answer 7

x⁴-5x³+11x²-25x+30 = 0      ← Check to see if +11x² can be broken up to form two
                                            trinomials that share a common factor. Consider
                                            x⁴-5x³ ≈ x²(x²-5x+C) ... For x²-5x+C to factor, the factors of C
                                            must add to -5 ... If we try -3 & -2, that would make C = +6
                                            so that 11x² would be broken up into +6x²+5x². And, notice
                                            that 5 is a factor of both -25 & +30. So, C=+6 looks good.
x⁴-5x³+6x² + 5x²-25x+30 = 0
x²(x²-5x+6) + 5(x²-5x+6) = 0      ← Now, factor out (x²-5x+6)
              (x²+5)(x²-5x+6) = 0      ← Now, factor (x²-5x+6)
             (x²+5)(x-3)(x-2) = 0

Roots are for:
  i. x²+5 = 0  ⇒  x = ±i√5
ii.   x-3 = 0  ⇒  x = 3
iii.   x-2 = 0  ⇒  x = 2

       ANSWER
x = 2, 3, i√5, -i√5

——————————————————————————————————————
     x³-x²-13x+13 = 0
  x³-x² − 13x+13 = 0
x²(x-1) - 13(x-1) = 0            ← Now, factor out (x-1)
       (x²-13)(x-1) = 0
Roots are for:
i. x²-13 = 0  ⇒  x = ±√(13)
ii.    x-1 = 0  ⇒  x = 1

       ANSWER
x = 1, √(13), -√(13) *eliminate 1*
****ANSWER IS 3*****

Quick example for ya in this type of math

all 3 roots of our polynomial equation of degree 3 are real. Since (x−3) is a factor, then x=3 is a root. Since (4x+1) is a factor, then x=−​4​​1​​ is a root. Since (x+2) is a factor, then x=−2 is a root.

x^3-x^2-13x+13 = (x-1)(x^2 - 13) = (x-1) (x-√13) (x + √13) -> roots are 1, √13, -√13

x^4-5x^3+11x^2-25x+30 = (x-2)(x^3-3x^2+5x-15) = (x-2) (x-3) (x^2 + 5) = (x-2) (x-3) (x - i√5) (x + i√5) -> roots are 2, 3, i√5, -i√5
=========================
Very simple tip I learned in early years of 8th grade math. Hope this helps :)
Have a nice day. Tag me on anymore you need help with.
(in conclusion answer is 3)

Answer 8

thanks. can you help me with aanother question

Answer 9

Find all the zeros of the equation.
-4x4 - 44x2 + 3600 = 0
A. 5, -5, 6i, -6i
B. 5, 6i
C. 5, -5, 6i, 0
D. -5, -6i

Answer 10

its not b

Answer 11

Well yes we know it's definitely not B, Let f(x) = x^4 – 6x^2 – 7x – 6
Note factors of the constant term –6 are ±1, ±2, ±3 and ±6
f(−2) = 0 means x+2 is a factor of f(x)
f(3) = 0 means x−3 is a factor of f(x)
Do long division or synthetic division or by inspection
f(x) = (x+2)(x−3)(x²+x+1)
(x+2)(x−3)(x²+x+1) = 0
x = −2 or x = 3 or x²+x+1 = 0
Solving the quadratic equation, in addition to the 2 real solutions,
x = ½(−1 ± i√3). Noticing the listing above I wrote, we know it's not C and D as well. What do you think the answer is? :)

Answer 12

a.

Answer 13

@caitlinnicole11 correct! :)

Answer 14

thanks