There is a single pair $(a,b)$ such that $x^2+2(1+a)x+(3a^2+4ab+4b^2+2)=0$. For this pair, what is $a+b$?
I know how to find rational roots now because of my last question asked, but what about real roots? What do I do?
There is a single pair $(a,b)$ such that $x^2+2(1+a)x+(3a^2+4ab+4b^2+2)=0$. For this pair, what is $a+b$?
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algebra-precalculus
roots
quadratics
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0Do you mean *for a given $x$* there is such a pair $(a,b)$, and we don't care to find what that $x$ is? – 2017-01-08
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0Yes $\ \ \ \ \ $ – 2017-01-08
2 Answers
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$$0=x^2+2(1+a)x+(3a^2+4ab+4b^2+2) =\\(x+1+a)^2+(3a^2+4ab+4b^2+2-1-2a-a^2) =\\(x+1+a)^2+(2a^2+4ab+4b^2+1-2a) =\\(x+1+a)^2+(a^2+4ab+4b^2)+(a^2-2a+1) \\ =(x+1+a)^2+(a+2b)^2+(a-1)^2 $$
Since $$(x+1+a)^2+(a+2b)^2+(a-1)^2=0$$ you get $$x+1+a=0\\ a+2b=0\\ a-1=0$$
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Hint: Completing squares, we have: $(x + a+1 )^2 + (a-1)^2 + (a+2b)^2 = 0$. Can you continue?