Question 985339: What are the real roots of this equations?
(3+x)2(x+7)(x-2)4=0
(x+8)(x-7)(x^2-2x+5)=0
x^2(x-9)(2x+1)=0
(x+4)(x^2-x+3)=0
(x+1)(x-3)=0
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! You set each parenthetical term equal to zero and solve.
for the first, (3+x)=0, x= -3; the others are -7, and 2.
for the second, it is -8, +7, and the solution to x^2-2x+5=0. That is done with the quadratic formula.
The result is x=(1/2) [2+/- sqrt (4-20)]=(1/2) [(2+/-sqrt (-16)]=1+/- 2i., because sqrt 16=4i and we take half.
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the third is 0,9,, and -(1/2)
the fourth is -4 and x^2-x+3 by the quadratic is (1/2) [1+/- sqrt (1-12)]= (1/2)+/- (1/2)sqrt (11 i)
;
The last: x=-1,3
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