Conditions On Roots Of Quadratics

Go back to  'Quadratic Equations'

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{\(\begin{align}&af\left( {{k_1}} \right) < 0\\&af\left( {{k_2}} \right) < 0\end{align}\)}\)

Learn from the best math teachers and top your exams

  • Live one on one classroom and doubt clearing
  • Practice worksheets in and after class for conceptual clarity
  • Personalized curriculum to keep up with school