In My Exam a question was as follows: $$$$Find all real values of $a$ such that the polynomial $$q(x)=(x+7)^2(x+2)^2+a$$ has exactly one double root. $$$$As we see that the polynomial $$p(x)=(x+7)^2(x+2)^2$$ has two double roots at $x=-7, -2$ and by calculating the derivative we get that this polynomial has a local maximum at the point $x=\frac{-9}{2}$. Now as by adding or subtracting any constant to or from a polynomial just vertically translates the graph of the polynomial. Now if $a > 0$ then the polynomial $q(x)>0$ and if $a=0$ then $q(x)=p(x)$ which has two double roots. Now as $p(x)$ has exactly two double roots and only one local Extremum so to get exactly one double root the graph of $p(x)$ must be translated vertically downwards by $p(\frac{-9}{2})$ which is a local maximum point. So we get $a=-p(\frac{-9}{2})=-\frac{625}{16}$. $$$$Was My Solution Correct??
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Oh sorry for that – user728159 Sep 20 '20 at 15:45
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How did your ISI exam go? – user600016 Sep 20 '20 at 15:45
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1My exam went nice – user728159 Sep 20 '20 at 15:50
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But can you plz tell wether my solution was correct or not?? – user728159 Sep 20 '20 at 15:50
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Yes, this is correct. A double root is a value for which $q(x)=0$, $q'(x)=0$ but $q''(x)\neq 0$. Since $q'(x)\equiv p'(x)$, the only possible places for double roots are the three roots of $p'(x)$ (none of which are roots of $p''(x)$), but two of these roots occur for the same value of $a$, meaning that the only place we can have exactly one double root is at $x=-9/2$, and as you say this occurs if and only if $a=-p(-9/2)$.
Especially Lime
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