Question

Can someone check my answer please??
x^3 + x^2=-3x-3
My answer: i√3, 1

Answer 1

You should have 3 roots as x^3 is the highest power in the equation. Also, when you have complex roots, they always come in conjugate pairs.

Answer 2

@Euler271 you entered the equation incorrectly. You did \(x^3+x^2-3x-3 = 0\), but it is actually \(x^3+x^2=-3x-3\) (which is the same as \( x^3+x^2+3x+3=0\))

Answer 3

x^3 +x^2 =-3x -3

factoriz out what you can on the left and right part

x^2(x+1)=-3(x+1) add to both sides 3(x+1)

x^2(x+1)+3(x+1)= 0

factoriz out (x+1)

(x^2 +3)(x+1)=0 so what mean that

x^2 +3=0
x^2 =-3

x_1,2=+/- sqrt(-3) = +/- isqrt3

and x+1=0

x=-1

than will be

x_1,2=+/- isqrt3
x_3 = -1

Answer 4

If you want that typeset a little neater:\[x^3 +x^2 =-3x -3 \]\[x^2(x+1)=-3(x+1)\]Add 3(x+1) to both sides\[x^2(x+1)+3(x+1)=0\]Factor out \((x+1)\)\[(x+1)(x^2+3)=0\]With a product = 0, solve each term = 0 to find all solutions:\[x+1=0\]\[x^2+3=0\]\[x=-1\]\[x=\pm i\sqrt{3}\]

Answer 5

Okay thanks everyone!

Answer 6

As I mentioned, if you have a polynomial with only real coefficients (no complex numbers), your complex roots will always come in conjugate pairs \((a\pm bi)\) so that when multiplied out, the \(i\) terms will become real numbers.

Answer 7

\[(a+bi)(a-bi) = a^2+abi - abi + b^2i^2 = a^2 + (-1)b^2 = a^2-b^2\]

Answer 8

Okay thanks for your help! @whpalmer4