Asked by Ariana
Solve polynomial equation by finding all real roots
X^3+ 6x^2+9x+2=0
I get that the answer is x=-2 when you do synthetic division and that the equation is (x+2)(x^2+4x+1), BUT
Where does the other answer . -2+/-√3 come from?
Than you!
X^3+ 6x^2+9x+2=0
I get that the answer is x=-2 when you do synthetic division and that the equation is (x+2)(x^2+4x+1), BUT
Where does the other answer . -2+/-√3 come from?
Than you!
Answers
Answered by
Reiny
going from your factored result,
(x+2)(x^2+4x+1) = 0
so x+2 = 0 or x^2 + 4x + 1 = 0
from x+2 = 0 -----> you get your root of x = -2
I you are studying equations at this level, surely you MUST know the
quadratic equation formula, then ...
x^2 + 4x + 1 = 0 , were a=1, b=4, and c=1
x = (-4 ± √(4^2 - 4(1)(1)) )/2
= ( -4 ± √12)/2
= (-4 ± 2√3)/2
divide each term by 2
= -2 ± √3
you could also complete the square ....
x^2 + 4x + 1 = 0
x^2 + 4x + ... = -1 + ....
x^2 + 4x + 4 = -1 + 4
(x+2)^2 = 3
x+2 = ±√3
x = -2 ±√3
(x+2)(x^2+4x+1) = 0
so x+2 = 0 or x^2 + 4x + 1 = 0
from x+2 = 0 -----> you get your root of x = -2
I you are studying equations at this level, surely you MUST know the
quadratic equation formula, then ...
x^2 + 4x + 1 = 0 , were a=1, b=4, and c=1
x = (-4 ± √(4^2 - 4(1)(1)) )/2
= ( -4 ± √12)/2
= (-4 ± 2√3)/2
divide each term by 2
= -2 ± √3
you could also complete the square ....
x^2 + 4x + 1 = 0
x^2 + 4x + ... = -1 + ....
x^2 + 4x + 4 = -1 + 4
(x+2)^2 = 3
x+2 = ±√3
x = -2 ±√3
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