Conditions On Roots Of Quadratics
Let \(f\left( x \right) = a{x^2} + bx + c\) and \(\alpha ,\beta \) be its roots. In this section, we deal with the problem of placing constraints on \(a, b, c\) given some constraint(s) on \(\alpha \,{\rm{and}}\,\beta \) .
In section-1, we saw the constraint \(D \ge 0\) given that \(\alpha ,\beta \) are real. Similarly, \(D < 0\) if \(\alpha ,\beta \) are non-real. This is what we mean by saying that the nature of \(\alpha ,\beta \) places a constraint on \(a, b, c:\)
\[\begin{align}&\alpha ,\beta \,{\text{real}} \qquad \quad\;\; \Rightarrow \quad {b^2} - 4ac \geqslant 0 \\ &\alpha,\beta \,{\text{non - real}} \quad \Rightarrow \quad {b^2} - 4ac < 0 \\ \end{align}\]
Now we will deal with more specific constraints:
(A) \(\fbox{Both roots are of the same sign}\)
Since the roots are real, \(D \ge 0.\) Since they are of the same sign, \(\alpha \beta > 0.\)
The graphs below illustrate examples for \({\text{a}}\,\, > \,\,0\)
Hence, the constraints are:
\(\fbox{\(\begin{align}&{b^2} - 4ac \ge 0\\\,\,\,\,\,\,\,\,\,&\qquad\;\;\frac{c}{a} > 0\end{align}\)}\)
(B) \(\fbox{Roots are of opposite sign}\)
For real roots, \(D \ge 0.\) For roots of opposite sign, \(\alpha \beta < 0.\) However, notice that \(\alpha \beta < 0\) ensures that \(D > 0\) and hence writing the first constraint is unnecessary (Why?).Therefore all we require is \(\alpha \beta < 0.\)
The graphs below illustrate this case.
The required constraint is:
\(\boxed{\frac{c}{a}< 0}\)
(C) \(\fbox{\({\rm{Both}}\,{\rm{roots}}\,{\rm{are}}\,{\rm{positive}}\)}\)
If you have followed the previous two cases properly, this case should be straight forward. We first of all require \(D \ge 0.\) Since both the roots are positive, \(\alpha \beta > 0\,{\rm{and}}\,\alpha + \beta > 0.\)
\(\fbox{\(\begin{align}&{b^2}\,\, - 4ac \ge \,0\\&\,\,\,\,\,\,\,\frac{{ - b}}{a} > \,0\\&\,\,\,\,\,\,\,\,\,\frac{c}{a} > \,0\end{align}\)}\)
(D) \(\fbox{\({\rm{Both}}\,{\rm{roots}}\,{\rm{are}}\,{\rm{negative}}\)}\)
This is similar to the previous case:
\(\fbox{\(\begin{align}{{b}^{2}}\,\,-4ac\ge \,0 \\ \,\,\,\,\,\,\,\frac{-b}{a}<\,0 \\ \,\,\,\,\,\,\,\,\,\frac{c}{a}>\,0 \\ \end{align}\)}\)
(E) \(\fbox{\({\rm{Roots}}\,{\rm{lie}}\,{\rm{on}}\,{\rm{either}}\,{\rm{side}}\,{\rm{of}}\,k\)}\)
This means that \(\alpha < k,\,\,\beta > k\)
We see that if \(a > 0,\,f\left( k \right) < 0\) and if \(a < 0,f\left( k \right) > 0\) or to put concisely, \(af\left( k \right) < 0.\) This represents a necessary and also sufficient constraint for our requirement (verify that this condition alone is sufficient):
\(\fbox{\(af\left( k \right) < 0\)}\)
(F) \(\fbox{\({\rm{Both}}\,{\rm{the}}\,{\rm{roots}}\,{\rm{are}}\,{\rm{less}}\,{\rm{than}}\,k\)}\)
We see that for the left hand graph \(\left( {a > 0} \right),f\left( k \right) > 0\) and for the right hand graph \(\left( {a < 0} \right),f\left( k \right) < 0.\) We can say this concisely as \(af\left( k \right) > 0.\) We also require \(D \ge 0\) and since both the roots are less than k, \(\alpha + \beta < 2k\)
Hence, the constraints are:
\(\fbox{\(\begin{align}&D \ge 0\\&af\left( k \right) > 0\\ &- \frac{b}{a} < 2k\end{align}\)}\)
(G) \(\fbox{\({\rm{Both}}\,{\rm{the}}\,{\rm{roots}}\,{\rm{are}}\,{\rm{greater}}\,{\rm{than}}\,k\)}\)
Here again, we see that \(af\left( k \right) > 0.\) Also, \(\alpha + \beta > 2k\)
\(\fbox{\(\begin{align}&D \ge 0\\&af\left( k \right) > 0\\ &- \frac{b}{a} > 2k\end{align}\)}\)
(H) \(\fbox{\({\rm{Both}}\,{\rm{the}}\,{\rm{roots}}\,{\rm{lie}}\,{\rm{between}}\,{k_1}\,{\rm{and}}\,{k_2}\)}\)
First of all, \(D \ge 0\) for real roots. Now, notice that the x-coordinate of the vertex of the parabola lies between \({k_1}\,\,and\,\,{k_2}\). Also, whether a is positive or negative, notice that \(af({k_1}){\rm{ }}\,\,and \,\,af({k_2})\) are positive.
The constraints are:
\(\fbox{\(\begin{align}&D \ge 0\\&af\left( {{k_1}} \right) > 0\\&af\left( {{k_2}} \right) > 0\\&{k_1} < - \frac{b}{{2a}} < {k_2}\end{align}\)}\)
(I) \(\fbox{\({\rm{Exactly}}\,{\rm{one}}\,{\rm{root}}\,{\rm{lies}}\,{\rm{between}}\,{k_1}\,{\rm{and}}\,{k_2}\)}\)
In all the four possible cases, \(f\left( {{k_1}} \right){\rm{and}}\,f\left( {{k_2}} \right)\) are of opposite sign. We can write this as \(f\left( {{k_1}} \right)f\left( {{k_2}} \right) < 0\) . Try to see that once we write this constraint, the basic constraint \(D \ge 0\) becomes redundant (Why? Because \(f (k_1)\) and \(f (k_2)\) can be of opposite sign only if the graph crosses the axis; this means that writing \(f\left( {{k_1}} \right) \cdot f\left( {{k_2}} \right) < 0\) automatically implies that \(f\left( x \right)\) will have real roots).
The required constraint is.
\(\fbox{\(f\left( {{k_1}} \right) \cdot f\left( {{k_2}} \right) < 0\)}\)
(J) \(\fbox{\({k_1}\,{\rm{and}}\,{k_2}\,{\rm{lie}}\,{\rm{between}}\,{\rm{the}}\,{\rm{roots}}\)}\)
Here, we see that for both cases \(af\left( {{k_1}} \right){\rm{and}}\,af\left( {{k_2}} \right)\) will be negative. Notice again that this represents a sufficient condition.
\(\fbox{\(\begin{align}&af\left( {{k_1}} \right) < 0\\&af\left( {{k_2}} \right) < 0\end{align}\)}\)
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