Question

Find the roots of x^3 – x^2 + 2 = 0

Answer 1

.....

Answer 2

um not really

Answer 3

yes

Answer 4

Try +/- 1 and +/- 2. Always check ro Rational Roots. If you manage to find one, the world is a better place!

Answer 5

so
x = –1 + 2i, x = –1, x = – 2i, x = 1

Answer 6

Really? Four roots from a Cubic Equation? Something fishy about that.

Answer 7

yeah

Answer 8

1.
Find the roots of x3 – x2 + 2 = 0.
A.

x = 1 + i, x = 1 – i, x = –1

B.

x = –1 + 2i, x = –1, x = – 2i, x = 1

C.

x = –1 + i, x = –1 – i, x = –1

D.

x = 1 + 2i, x = 1 – 2i, x = 1

Answer 9

see

Answer 10

yes

Answer 11

Okay, then discard B immedaitely. It absolutely cannto be correct.

Now, quit guessing. Did you yet find a factoring? A root? Prove it.

Answer 12

x=1

Answer 13

x=-1

Answer 14

it means x-1=0

Answer 15

now plug x=-1

Answer 16

if it gives u ...zero ...then it is ur root

Answer 17

i mean first root

Answer 18

......
aye,
eh,
thank but
im thinking its C

Answer 19

Thinking? That's just not good enough. Prove it!!

x^3 – x^2 + 2 = 0

Rational Roots can be ONLY +/- 1, +/- 2

A quick inspection says there MUST be a Negative Real Root. Maybe it's rational!

Let's try x = -1

(-1)^3 - (-1)^2 + 2 = -1 -1 + 2 = 0 -- That's it!!!! x = -1 is a root. (x+1) is a factor.

Now we reduce the problem

(x+1)(x^2 - 2x + 2) = 0

That last part doesn't factor. Use the Quadratic Formula to find the other TWO roots.

Quadratic ==> Two
Cubic ==> Three
Quartic ==> Four
Quintic ==> Five

Answer 20

guyes theres more than one answer

Answer 21

A.

x = 1 + i, x = 1 – i, x = –1

B.

x = –1 + 2i, x = –1, x = – 2i, x = 1

C.

x = –1 + i, x = –1 – i, x = –1

D.

x = 1 + 2i, x = 1 – 2i, x = 1

see????

Answer 22

x^3-x^2+2 = 0
Factor the left hand side.
The left hand side factors into a product with two terms:
(x+1) (x^2-2 x+2) = 0
Solve each term in the product separately.
Split into two equations:
x+1 = 0 or x^2-2 x+2 = 0
Look at the first equation: Solve for x.
Subtract 1 from both sides:
x = -1 or x^2-2 x+2 = 0
Look at the second equation: Solve the quadratic equation by completing the square.
Subtract 2 from both sides:
x = -1 or x^2-2 x = -2
Take one half of the coefficient of x and square it, then add it to both sides.
Add 1 to both sides:
x = -1 or x^2-2 x+1 = -1
Factor the left hand side.
Write the left hand side as a square:
x = -1 or (x-1)^2 = -1
Eliminate the exponent on the left hand side.
Take the square root of both sides:
x = -1 or x-1 = i or x-1 = -i
Look at the second equation: Solve for x.
Add 1 to both sides:
x = -1 or x = 1+i or x-1 = -i
Look at the third equation: Solve for x.
Add 1 to both sides:
Answer:
x = -1 or x = 1+i or x = 1-i

Answer 23

x=0, 1 and 1/2+root7/2 i

Answer 24

thats not an optoin!

Answer 25

B. No good. There are four. There can be only 3.
D. No good. Does not include x = -1

@selenamalter You have shown NO work of your own. You cannot solve ANY problem with that kind of effort. Don't GUESS at the answer. Prove it!

Answer 26

(x+1) (x^2-2x+2)=0 solve it

Answer 27

x=-1 ...for second part...use quardatic eqaution

Answer 28

x=-1, x= 2+i/2
x=2-i/2

Answer 29

yeah u can get that

Answer 30

do u rem the formula for quardatic equation

Answer 31

it comes like 2+0r - square root of 4 i over 2

Answer 32

do it u will get that answer

Answer 33

Seriously, @basa, quit typing!
Serously, @selenamalter, show your work. What is your plan for a solution?

Answer 34

He has none. He's just guessing.

Answer 35

But I think that was obvious.

Answer 36

...... ya confusing meh...
oh um

Answer 37

ugh! forget it i quit

Answer 38

This will almost always be the result when you do no work of your own.

Answer 39

STHAP IT
I AM
eh, mabe i eh i dont undertand hmmm?

Answer 40

yeh only possible answes are x=-1,x=1+i,x=1-i

Answer 41

you better plug those values in the equations and check them out..it should give u zero..n it does

Answer 42

@ selenamalter I believe that you do have some clue. You have shown it on other problems in other threads. You just didn't show any of it in this thread. Not anything. Go back and look. There is nothing. That will never be good enough.

Answer 43

yep total of real and complex solutions as following:

Real: x=-1
Imag: x=1-1i,1+1i

Answer 44

x=-1.....it makes that equation zero so it is one of the roots of that eqations..do same for other two

Answer 45

x^3+x^2-2x^2-2x+2x+2=0 imples x^2(x+1)-2x(x+1)+2(x+1)=0 implies (x+1)(x^2-2x+1)=0