Quadratic Roots and Real Numbers

The Core Logic: Three Ingredients

Almost every location of roots problem can be solved by checking up to three key aspects of the quadratic function $f(x)=a{x}^2+bx+c$. Understanding these three ingredients is far more effective than trying to memorize six different cases.

1. Discriminant ($D=b^2-4ac$): This confirms if real roots exist in the first place. For most problems, we need real roots, so the condition is $D\ge 0$.
2. Value of the function at a point k ($f(k)$): The sign of $f(k)$ tells us whether the point k is "under the arch" of the parabola (between the roots) or "outside the arch." We typically use the combined condition $a\cdot f(k)$ to handle both upward ($a>0$) and downward ($a<0$) facing parabolas with a single rule.
3. Position of the Vertex ($x=-b/2a$): This tells us on which side of a point k the bulk of the parabola lies. It's the deciding factor when we need to place both roots to one side of k.

Let's see how these ingredients combine to solve different types of problems.

Case 1: A Number k Lies Between the Roots ($\alpha <k<\beta$)

This is the simplest scenario. If a number k is between the roots, the graph of the parabola must cross the x-axis on either side of k.

• If $a>0$ (upward parabola), the value $f(k)$ must be negative.
• If $a<0$ (downward parabola), the value $f(k)$ must be positive.

Notice that in both situations, $a$ and $f(k)$ have opposite signs. This gives us a single, powerful condition.

Condition: $a\cdot f(k)<0$

That's it. You might wonder, "What about $D\ge 0$?" If $a\cdot f(k)<0$, it means the parabola opens in one direction (e.g., up) but has a value on the opposite side of the x-axis (e.g., negative). To get back to the x-axis to form roots, it must cross it. Therefore, the existence of real roots is automatically guaranteed.

The following video segment explains this with excellent visualizations.

A common application is when a problem states that the roots are of opposite signs. This is simply Case 1 with $k=0$.

Case 2: Both Roots are Greater than a Number k ($k<\alpha \le \beta$)

This is the most fundamental case and requires all three of our ingredients to ensure the roots are not only real but also on the correct side of k.

Let's build the conditions logically:

1. Roots must be real: $D\ge 0$. If the problem specifies "distinct roots," use $D>0$.
2. k must be outside the roots: For k to be to the left of both roots, its functional value $f(k)$ must have the same sign as $a$. This prevents k from being between the roots. So, $a\cdot f(k)>0$.
3. The parabola must be on the correct side of k: Conditions 1 and 2 are also true if both roots are less than k. To distinguish, we must pin down the location of the parabola's vertex. The vertex must be to the right of k. So, $-b/2a>k$.

Conditions:

1. $D\ge 0$
2. $a\cdot f(k)>0$
3. $-b/2a>k$

Important Correction: The image Location of Roots - Both roots greater than k incorrectly states the vertex condition as k > -b/2a. As you can see from the graph, the vertex at x = -b/2a is to the right of k, so the correct condition is $-b/2a>k$. Always rely on the graphical logic.

The next video segment walks through this exact logic.

Variations on the Core Cases

All other scenarios are clever combinations or variations of the first two cases.

Case 3: Both Roots are Less than k ($\alpha \le \beta <k$)
The logic is nearly identical to Case 2. The only change is the vertex must now be to the left of k.

1. $D\ge 0$
2. $a\cdot f(k)>0$
3. $-b/2a<k$

Case 4: Both Roots Lie Between k1 and k2 ($k_1<\alpha \le \beta <k_2$)
This is just a combination of "both roots greater than k1" AND "both roots less than k2".

1. $D\ge 0$
2. $a\cdot f(k_1)>0$
3. $a\cdot f(k_2)>0$
4. $k_1<-b/2a<k_2$

Case 5: Exactly One Root Lies in $(k_1,k_2)$
This means one root is inside the interval and one is outside. Looking at the graph, this requires $f(k_1)$ and $f(k_2)$ to have opposite signs.
Condition: $f(k_1)\cdot f(k_2)<0$
Like Case 1, this single condition is sufficient. The existence of a root between $k_1$ and $k_2$ is guaranteed by the sign change.

The following video from MathonGo provides a concise walkthrough of these different configurations, which will help solidify your understanding.

Practice Problems

Your preference is for a practice-heavy approach, so let's apply this toolkit. The following resource contains JEE-level problems with solutions.

For further practice, the jeeprepguide resource has a set of problems. Attempting these will build your speed and confidence.

Conclusion

You've now covered one of the most important applications of quadratic theory for JEE Advanced. The key is not to blindly memorize formulas, but to use the graphical interpretation to construct the necessary conditions.

Key Takeaways:

• The Three Ingredients: Almost all problems can be solved by analyzing the Discriminant ($D$), the function's value at boundary points ($a\cdot f(k)$), and the vertex's position ($-b/2a$).
• Simple Cases, Simple Conditions: For problems where a number k lies between the roots, or when exactly one root is in an interval $(k_1,k_2)$, a single condition involving the product of function values is often sufficient ($a\cdot f(k)<0$ or $f(k_1)\cdot f(k_2)<0$). This is a powerful shortcut.
• Comprehensive Cases, Full Toolkit: For problems where both roots must be on one side of k or within an interval $(k_1,k_2)$, you need to apply the full set of conditions to constrain the parabola completely.

In our next lesson, we will tackle another common problem type: finding conditions for two different quadratic equations to have one or both roots in common.