$$x^2-2(3m-1)x+2m+3=0$$ Find the sum of solutions. It says that the sum equals to $-1$. I just can't wrap my head around this? Any help? Thx
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is an $x$ missing? – Dr. Sonnhard Graubner Jun 29 '17 at 12:21
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@Dr.SonnhardGraubner yes. Sorry – kenobe Jun 29 '17 at 12:21
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In a quadratic equation $ax^2+bx+c=0$ the sum of the solutions (if any) is equal to $-b/a$. In fact, if $x_1$ and $x_2$ are the solutions then $$ax^2+bx+c=a(x-x_1)(x-x_2)=ax^2-a(x_1+x_2)x+a(x_1\cdot x_2)$$ and by comparing the coefficients we get $x_1+x_2=-b/a$.
In your case $x_1+x_2=2(3m-1)$. It seems that the sum of the (complex) solutions is $-1$ if $m=1/6$.
Robert Z
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$$x_{1,2}=3m-1\pm\sqrt{9m^2-8m-2}$$ therefore $$x_1+x_2=2(3m-1)$$
Dr. Sonnhard Graubner
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Yes that's exactly what I was thinking but in the "solution" area they are using absolute value brackets. To be more precise it states : |x-2|+|x+3|=7. – kenobe Jun 29 '17 at 12:28
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i don't know what this equation means, can you post the whole text please? – Dr. Sonnhard Graubner Jun 29 '17 at 12:30
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Yes sir. The question is : find the sum of all solutions of the given equation. (The equation from above) – kenobe Jun 29 '17 at 12:32
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Okay. But I find it weird that when i plug -1 into 2(3m-1) I do get m=1/6 so for m=1/6 the solution is correct – kenobe Jun 29 '17 at 12:44