Quadratic Equations and Inequalities

Why This Topic Matters for NDA

Quadratic equations sit at the heart of NDA algebra. They connect to sequences, coordinate geometry, trigonometry and even physics-style word problems. UPSC examiners love this chapter because a single idea — the roots of ax² + bx + c = 0 — can be tested in many disguises.

Most questions are short and scoring if you know the standard results. You rarely need to fully solve an equation; instead you compare coefficients, use the discriminant, or read off the sum and product of roots. A student who has truly mastered this chapter can often look at four answer options and eliminate three of them before doing any real calculation.

The reason this chapter rewards smart preparation is that almost every question is built on the same small set of ideas: the standard form, the roots, the discriminant, and the two coefficient relations. Once these four ideas are second nature, the chapter stops feeling like algebra and starts feeling like pattern recognition. That is exactly the mindset The Cavalier trains its students to develop for the NDA written paper.

Standard Form and Basic Vocabulary

Every quadratic can be arranged into the standard form:

The values of x that satisfy the equation are called its roots or solutions. A quadratic always has exactly two roots (counting repeated and complex roots), because it is a degree-2 polynomial. This is a special case of the Fundamental Theorem of Algebra, which says a polynomial of degree n has exactly n roots in the complex number system.

It helps to picture the graph of y = ax² + bx + c, which is a parabola. The roots of the equation are simply the points where this parabola crosses the x-axis, because there y equals zero. Connecting the algebra to the graph makes the whole chapter far easier to remember.

• If both roots are real and different, the curve y = ax² + bx + c cuts the x-axis at two distinct points.
• If the roots are equal, the parabola just touches the x-axis at one point (the vertex sits on the axis).
• If the roots are complex, the parabola never meets the x-axis at all; it floats entirely above or below it.

The sign of a decides which way the parabola opens: a > 0 opens upward like a valley, while a < 0 opens downward like a hill. Keeping this picture in mind will save you in inequality questions later.

The Quadratic Formula

The most reliable way to find the roots is the quadratic formula, obtained by completing the square on the standard form.

The quantity under the root sign, b² − 4ac, is so important it gets its own name — the discriminant, written as D or Δ. It alone decides what kind of roots you get, without you having to compute them.

The formula itself comes from a neat trick called completing the square. Starting from ax² + bx + c = 0, you divide through by a, move the constant across, and add a square term to both sides to fold the left side into a perfect square. The final rearrangement gives exactly the formula above. You do not need to reproduce this derivation in the exam, but knowing where the formula comes from helps you trust it and remember the −b and the 2a in the right places.

Discriminant and Nature of Roots

The discriminant D = b² − 4ac tells you the nature of the roots at a glance. Memorise this table; it answers a large share of NDA questions directly.

Two extra refinements often appear in the exam:

• If a, b, c are rational and D is a perfect square, the roots are rational; otherwise they are irrational and come in conjugate pairs like p ± √q.
• Complex roots of a real quadratic always occur as conjugates: if α + iβ is a root, so is α − iβ.

Here is how to use this in practice. Suppose a question gives you 3x² + 5x + 7 = 0 and asks for the nature of the roots. You compute D = 5² − 4(3)(7) = 25 − 84 = −59. Since D is negative, you can confidently mark “complex / imaginary roots” without solving anything further. That is the kind of ten-second answer the discriminant gives you.

Sum and Product of Roots

If α and β are the roots of ax² + bx + c = 0, then two beautiful relations connect them to the coefficients — no need to actually find the roots.

These let you build an equation from its roots. A quadratic with roots α and β can be written as:

Useful derived results that NDA examiners test:

• α² + β² = (α + β)² − 2αβ
• (α − β)² = (α + β)² − 4αβ
• 1÷α + 1÷β = (α + β) ÷ αβ
• α³ + β³ = (α + β)³ − 3αβ(α + β)

Why do these relations exist? If α and β are the roots, then ax² + bx + c = a(x − α)(x − β). Expanding the right side and matching coefficients with the left immediately gives α + β = −b÷a and αβ = c÷a. Understanding this one expansion means you will never confuse which relation carries the minus sign.

Forming Equations and Special Root Conditions

Many NDA questions hide simple conditions inside words. Learn to translate them:

• Roots are equal → D = 0.
• Roots are reciprocal (β = 1÷α) → product = 1, so c = a.
• Roots are equal in magnitude but opposite in sign → sum = 0, so b = 0.
• One root is zero → product = 0, so c = 0.
• Both roots positive → D ≥ 0, sum > 0 and product > 0.

Worked Example

Let us combine the discriminant and sum–product ideas in one typical NDA-style problem.

Inequalities: The Basics

An inequality compares two expressions using <, ≤, > or ≥. Solving one means finding all values of x that make it true. The rules are like equations, with one crucial twist.

For a linear inequality like 3x − 5 ≤ 7, just isolate x: add 5 to both sides to get 3x ≤ 12, then divide by the positive number 3 to get x ≤ 4. The answer is an interval, written x ≤ 4 or ( −∞, 4 ]. Notice we did not flip the sign because we divided by a positive number.

It is worth getting comfortable with interval notation, because NDA options are often written that way. A square bracket [ or ] means the endpoint is included (used with ≤ and ≥), while a round bracket ( or ) means the endpoint is excluded (used with strict < and >). Infinity always takes a round bracket because it is never actually reached.

Solving Quadratic Inequalities

To solve a quadratic inequality such as ax² + bx + c > 0, follow the sign method:

1. Move everything to one side so it compares with 0.
2. Factorise to find the critical points (the roots).
3. These roots split the number line into intervals.
4. Test the sign of the expression in each interval and pick the ones that satisfy the inequality.

A quick memory hook: with a positive leading coefficient, the parabola opens upward, so it is negative in the dip between the roots and positive outside. If the leading coefficient is negative, the picture flips, so it pays to first make a positive (multiply the whole inequality by −1, remembering to flip the sign) before applying the rule.

Let us see the method on x² − x − 6 < 0. Factorising gives (x − 3)(x + 2) < 0, so the critical points are x = 3 and x = −2. Since a = 1 is positive and we want the expression to be less than zero, the answer is the region between the roots: −2 < x < 3. Always write the smaller root on the left so the interval reads correctly.

Common Mistakes to Avoid

A handful of slips cost students easy marks every year. Watch for these:

• Forgetting a ≠ 0. If the coefficient of x² can be zero, treat that case separately.
• Sign errors in the formula. The numerator is −b, so for b = −7 it becomes +7.
• Mixing up sum and product. Sum is −b÷a (note the minus), product is c÷a (no minus).
• Not flipping the inequality when dividing by a negative.
• Including wrong endpoints. Use closed brackets only for ≤ and ≥, open for strict < and >.

Previous-Year Style Question

This question blends the discriminant with a coefficient condition — a classic NDA pattern.

Quick Revision