Why the graph beats the algebra
Take the question: for what values of k does x² − kx + 4 = 0 have both roots greater than 1? Solving for the roots gives an expression with a square root and two inequalities to manage simultaneously. Thinking graphically instead gives three conditions immediately - the parabola must actually cross the axis (D ≥ 0), it must be positive at x = 1 since both roots lie to the right of it, and its vertex must lie to the right of 1.
Those three conditions are mechanical, and they replace all the case analysis. The same triple applies to every location question with the point and the direction adjusted, which is why this chapter rewards learning one method rather than many special cases.
How JEE actually asks Quadratic Equations
NTA publishes no chapter-wise weightage, so figures circulating online are coaching estimates from past papers; confirm your session's syllabus and pattern in the official NTA information bulletin.
Past papers show that direct root-finding is rare. More common are questions using the relations between roots and coefficients to evaluate a symmetric expression, questions about the nature or location of roots as a parameter varies, and quadratic inequalities. The chapter also feeds Complex Numbers, since roots with negative discriminant are conjugate pairs.
Key concepts, compressed
- ✓For ax² + bx + c = 0 with a ≠ 0, the roots are (−b ± √D)/2a where D = b² − 4ac.
- ✓D > 0 gives real distinct roots, D = 0 gives equal roots, D < 0 gives a conjugate pair of complex roots.
- ✓The sum of the roots is −b/a and their product is c/a, which lets symmetric expressions be evaluated without solving.
- ✓The graph is a parabola opening upward when a > 0 and downward when a < 0, with vertex at x = −b/2a.
- ✓For real coefficients, complex roots occur in conjugate pairs; for rational coefficients, irrational roots occur in conjugate surd pairs.
- ✓A quadratic inequality is solved by factorising and testing the sign of each interval, never by cross-multiplication.
Results you need before attempting the questions
| Quantity | Formula | Note |
|---|---|---|
| Roots | x = (−b ± √D)/2a | requires a ≠ 0 |
| Discriminant | D = b² − 4ac | |
| Sum of roots | α + β = −b/a | |
| Product of roots | αβ = c/a | |
| Difference of roots | |α − β| = √D/|a| | |
| Equation from roots | x² − (sum)x + product = 0 | |
| α² + β² | (α + β)² − 2αβ | |
| α³ + β³ | (α + β)³ − 3αβ(α + β) | |
| 1/α + 1/β | (α + β)/αβ | |
| Vertex | x = −b/2a | min if a > 0 |
| Both roots > k | D ≥ 0, a·f(k) > 0, −b/2a > k | |
| Roots of opposite sign | product < 0 | D > 0 follows |
The five mistakes that cost the most marks
- ✓Forgetting to check a ≠ 0 when the leading coefficient contains a parameter. If it can vanish, the equation becomes linear with one root, and that case is usually the point of the question.
- ✓Cross-multiplying an inequality by an expression whose sign is unknown. Multiplying by a negative quantity flips the inequality, so factorise and use a sign chart instead.
- ✓Treating 'both roots positive' as requiring only a positive product. You need D ≥ 0, a positive sum and a positive product together.
- ✓Ignoring the sign of a when discussing maxima and minima. The parabola opens downward for a < 0, so the vertex is a maximum, not a minimum.
- ✓Assuming complex roots always come in conjugate pairs. That holds only when the coefficients are real - with complex coefficients it fails.
Practice set 1: roots and the discriminant
1. Solve x² − 5x + 6 = 0.
x = 2 and x = 3. The expression factorises as (x − 2)(x − 3), so the roots are 2 and 3. Factorising is faster than the formula whenever the numbers are friendly, and you can spot it by looking for two numbers whose sum is 5 and whose product is 6 - which is exactly the roots-and-coefficients relation in disguise.
2. What is the discriminant of that equation, and what does it tell you?
D = 1, indicating two real and distinct roots. D = b² − 4ac = 25 − 24 = 1. Because D is positive the roots are real and different, and because it is a perfect square with rational coefficients they are also rational - which is why the factorisation in question 1 worked so cleanly.
3. What can you say about the roots of x² − 4x + 4 = 0?
They are real and equal, both x = 2. Here D = 16 − 16 = 0, and the expression is the perfect square (x − 2)². Geometrically the parabola touches the x-axis at exactly one point rather than crossing it - the vertex sits on the axis.
4. What about x² + x + 1 = 0?
The roots are complex conjugates. D = 1 − 4 = −3, which is negative, so there are no real roots and the parabola never meets the x-axis. Since the coefficients are real, the two roots are conjugates of each other. These particular roots are the non-real cube roots of unity, which links directly to the Complex Numbers chapter.
5. Without solving, find the sum and product of the roots of x² − 5x + 6 = 0.
Sum 5, product 6. Sum = −b/a = −(−5)/1 = 5 and product = c/a = 6/1 = 6. These match the roots 2 and 3 found earlier, as they must. The value of these relations is that they work when the roots are ugly or complex, which is when you would least want to compute them.
6. Form the quadratic equation whose roots are 2 and 3.
x² − 5x + 6 = 0. Use x² − (sum)x + product = 0 with sum 5 and product 6. Note the minus sign before the sum - writing x² + 5x + 6 gives roots −2 and −3 instead, which is the most common slip in this question type.
7. Find the absolute difference between the roots of x² − 5x + 6 = 0.
1. Use |α − β| = √D/|a| = √1/1 = 1, which matches |3 − 2| = 1. This formula is useful when the roots are irrational, since it gives the spread without computing either root - and it follows from (α − β)² = (α + β)² − 4αβ.
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Practice set 2: symmetric expressions in the roots
8. For x² − 5x + 6 = 0, find α² + β².
13. Use the identity α² + β² = (α + β)² − 2αβ = 25 − 12 = 13. Check against the actual roots: 4 + 9 = 13. The identity matters because it works for any quadratic, including one whose roots are irrational or complex, without ever finding them.
9. Find 1/α + 1/β for the same equation.
5/6. Combine into a single fraction: 1/α + 1/β = (α + β)/αβ = 5/6. Verifying directly, 1/2 + 1/3 = 5/6. Almost every symmetric expression reduces to the sum and product this way, which is why those two quantities are worth extracting first.
10. Find α³ + β³ for the same equation.
35. Use α³ + β³ = (α + β)³ − 3αβ(α + β) = 125 − 3(6)(5) = 125 − 90 = 35. Checking: 8 + 27 = 35. Memorise this identity alongside the one in question 8 - between them they cover the large majority of symmetric-expression questions.
11. Form the equation whose roots are double those of x² − 5x + 6 = 0.
x² − 10x + 24 = 0. Doubling the roots gives 4 and 6, so the new sum is 10 and the new product is 24. In general, doubling both roots doubles the sum and quadruples the product - and you can also get there by substituting x/2 for x in the original equation, which is the systematic method.
12. Form the equation whose roots are the reciprocals of those roots.
6x² − 5x + 1 = 0. The reciprocals are 1/2 and 1/3, giving sum 5/6 and product 1/6, so x² − (5/6)x + 1/6 = 0, which clears to 6x² − 5x + 1 = 0. The shortcut worth knowing: reversing the coefficients of a quadratic produces the equation with reciprocal roots.
13. What is the condition for two quadratics to have a common root?
Subtract one equation from the other to eliminate the x² term, giving a linear equation whose solution is the common root - then substitute back to confirm it satisfies both. If both roots are common, the two equations are proportional: a₁/a₂ = b₁/b₂ = c₁/c₂.
Practice set 3: nature and location of roots
14. What conditions ensure both roots of ax² + bx + c = 0 are positive?
D ≥ 0, sum of roots positive, and product of roots positive - all three together. A positive product alone only guarantees the roots share a sign, which could be negative; a positive sum alone does not prevent one root being negative and larger in magnitude. Questions offer each condition individually as a distractor.
15. What conditions ensure both roots are negative?
D ≥ 0, sum of roots negative, and product of roots positive. The product is positive in both this case and question 14 - it only tells you the signs match. The sum is what distinguishes them, which is why relying on the product alone fails.
16. What single condition ensures the roots have opposite signs?
The product of the roots is negative, that is c/a < 0. This automatically forces D > 0, since real distinct roots are required for opposite signs to be meaningful - so no separate discriminant check is needed. This is the one location condition that reduces to a single inequality.
17. What three checks answer a 'both roots greater than k' question?
First, D ≥ 0 so the roots are real. Second, a·f(k) > 0, meaning the parabola takes the same sign at k as its leading coefficient - which places k outside the interval between the roots. Third, the vertex −b/2a > k, which places that interval to the right of k. Together these are necessary and sufficient, and they replace all case analysis.
18. Why must you check that a ≠ 0 when the leading coefficient contains a parameter?
Because the equation stops being quadratic when a = 0. For (k − 1)x² + 2x + 3 = 0, the value k = 1 gives a linear equation with a single root, which is usually a case the question intends you to identify separately. Answers that assume two roots throughout miss it entirely.
19. How does sketching the parabola help with root questions?
It converts algebraic conditions into visible ones. The sign of a gives the opening direction, D gives the number of axis crossings, f(k) tells you which side of the curve the point k lies on, and the vertex locates the turning point. Almost every location question is answered by reading those four facts off a rough sketch.
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Practice set 4: inequalities and extrema
20. Solve x² − 5x + 6 > 0.
x < 2 or x > 3. Factorise as (x − 2)(x − 3) > 0. The expression is positive when both factors share a sign, which happens outside the roots. Since the parabola opens upward, it lies above the axis outside the interval between its roots - a picture that answers the question without any sign chart.
21. Solve x² − 5x + 6 < 0.
2 < x < 3. This is the complement of the previous answer, excluding the roots themselves. An upward parabola is below the axis strictly between its roots. Note the strict inequalities: at x = 2 and x = 3 the expression equals zero, which satisfies neither a strict less-than nor a strict greater-than.
22. Why should you never cross-multiply when solving an inequality?
Because multiplying by an expression of unknown sign may flip the inequality without warning. From (x − 1)/(x − 2) > 0 you cannot multiply by (x − 2) unless you know its sign. The safe method is to bring everything to one side, factorise, and use a sign chart across the intervals defined by the zeros.
23. Find the minimum value of x² − 4x + 5.
1, attained at x = 2. Complete the square: x² − 4x + 5 = (x − 2)² + 1. Since a square is never negative, the smallest possible value is 1 when x = 2. Equivalently the vertex is at x = −b/2a = 4/2 = 2, and substituting gives 4 − 8 + 5 = 1.
24. Where is the vertex of a parabola, and when is it a maximum?
At x = −b/2a. It is a minimum when a > 0, since the parabola opens upward, and a maximum when a < 0. Forgetting to check the sign of a is why students sometimes report a minimum for a downward-opening parabola - the vertex formula is the same either way, but its meaning is not.
25. What is the range of f(x) = x² − 4x + 5 over all real x?
All values greater than or equal to 1, so [1, ∞). The vertex value is the minimum and the parabola opens upward, so it takes every value above it and none below. For a downward-opening parabola the range would instead be everything up to and including the vertex value.
Practice set 5: conjugate roots and parameters
26. If a quadratic with real coefficients has one complex root, what is the other?
Its complex conjugate. Because the coefficients are real, the discriminant's square root contributes ±i√|D| symmetrically, so the roots are p + qi and p − qi. Note the condition carefully: with complex coefficients allowed, this no longer holds, and questions test exactly that qualification.
27. If a quadratic with rational coefficients has one root 2 + √3, what is the other?
2 − √3. Irrational roots occur in conjugate surd pairs when the coefficients are rational, for the same structural reason as complex conjugates - the surd enters only through ±√D. If the coefficients were merely real rather than rational, this would not follow.
28. Form the quadratic equation with rational coefficients having 2 + √3 as a root.
x² − 4x + 1 = 0. The other root must be 2 − √3, so the sum is 4 and the product is (2 + √3)(2 − √3) = 4 − 3 = 1. Then x² − 4x + 1 = 0. Notice how the surd disappears from the product - which is precisely why the equation can have rational coefficients at all.
29. For what values of k does x² + kx + 4 = 0 have equal roots?
k = 4 or k = −4. Equal roots require D = 0, so k² − 16 = 0 and k = ±4. Both values are valid and omitting the negative one is the standard error. At k = 4 the repeated root is −2, and at k = −4 it is +2.
30. For what values of k does x² + kx + 4 = 0 have real roots?
k ≤ −4 or k ≥ 4. Real roots need D ≥ 0, so k² ≥ 16, giving |k| ≥ 4. Compare with question 29: equal roots occur at the two boundary values, and strictly between −4 and 4 the roots are complex. Solving k² ≥ 16 as k ≥ 4 alone, forgetting the negative branch, is the error this pairing exposes.
How to study this chapter efficiently
- ✓Extract the sum and product of the roots before doing anything else. Most questions never need the roots themselves.
- ✓Memorise two identities - α² + β² and α³ + β³ in terms of sum and product. They cover most symmetric expressions.
- ✓For any location question, draw the parabola and apply the three checks: D, the sign of a·f(k), and the vertex position.
- ✓Never cross-multiply an inequality. Factorise, find the zeros, and test the sign of each interval.
- ✓Whenever a parameter appears in the leading coefficient, write down the case where it vanishes as a separate line.
- ✓Practise the paired questions - equal roots versus real roots, both positive versus both negative. JEE sets them together precisely because the conditions differ by one inequality.
Turn this into active practice
The failure mode here is not computational. Students who can solve any quadratic still lose marks by giving only the positive branch of |k| ≥ 4, or by concluding both roots are positive from a positive product alone. Those are recognition errors, and they persist until a question exposes them.
The JEE Quadratic Equations quiz on QUFF generates fresh questions across the discriminant, root relations, location conditions, inequalities and extrema, marks them instantly and explains each answer. Do a mixed set and pay attention to whether your errors are arithmetic or conditional - the second kind is far more common and far more fixable.