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D.
Factorising the first quadratic gives x=2 and x=1, the second quadratic gives x=1,-2. For common value x=1, x+3 equals 4.
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x^2-3x+2=0 and x^3+2x^2-x-2=0
A) 1
B) 2
C) 3
D) 4
E) 5

x^2-3x+2=0
which is x^2-2x-x+2=0
which is x(x-2)-1(x-2)=0
which means (x-1) (x-2)=0, which means 1 and 2 are roots of the equation

Now x^3+2x^2-x-2=0
which is x^2(x+2)-1(x+2)=0
which is (x^2-1)(x+2)=0
which is (x-1)(x+1)(x+2)=0. ..(as a^2-b^2 = (a-b)(a+b))
so x can be either 1, -1 or -2 i.e 1, -1 and -2 are roots of the cubic equation

Common root between the two equation(quadratic and cubic) is only 1 at which both equation becomes zero, hence x = 1
hence x+3= 4 is the answer
D is the answer
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Below is my attempt

\(x^2\) - 3x + 2 = 0 --> (x-2)(x-1) = 0 --> x =2 or x = 1
\(x^3\) + 2\(x^2\)-x-2 = 0 --> (x+2)(\(x^2\)-1) = 0 --> x = 1 or x = -1 or x = -2

Combine 2 solutions --> x = 1
--> x +3 = 4

Choice D is the answer

Thank you!
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