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Quadratic Equations — Class 10 Maths Notes & Practice

July 31, 2026
Quadratic Equations — Class 10 Maths Notes & Practice

Quadratic equations are where Class 10 algebra starts to feel grown-up: suddenly there are two answers instead of one, and a formula with a square root inside it. Here is the reassuring truth — this chapter has only four real skills in it, and each one is a short, repeatable procedure. Once you can recognise a quadratic, factorise it, apply the formula, and read the discriminant, everything else in the chapter is a variation on those four moves.

🌐 इस अध्याय को हिंदी में पढ़ें →

We are going to build each skill from the ground up, with worked examples that start easy and climb to board level. Take your time on the factorisation section in particular — students who are comfortable there find the rest of the chapter almost restful.

🎯 Try This
Pick any rectangular object at home (a door, a table top, a rug) and write a quadratic equation for its area, using length = width + 3; solve it by factorisation and check the answer against the real measurements. (15-20 min)

Your Game Plan for This Chapter

  1. Make sure you can put any equation into standard form before anything else. Half the errors in this chapter happen here.
  2. Spend a full sitting on factorisation. Do fifteen of them, not five.
  3. Learn the quadratic formula by writing it out five times from memory — then use it on equations you have already factorised, so you can check yourself.
  4. Only then do discriminant questions; they are quick marks once the formula is solid.
  5. Finish with word problems, which combine everything and are worth the most marks.
  6. Do the worksheet at the end with a timer, as if it were a class test.

Study Notes

1. What Makes an Equation Quadratic

An equation is quadratic when, after all the tidying up is done, the highest power of the variable is exactly 2. Written neatly, it looks like this:

Key Idea — The standard form is ax² + bx + c = 0, where a, b and c are real numbers and a is not zero. If a were zero the x² term would vanish and you would be left with a linear equation.

The phrase “after all the tidying up is done” is doing a lot of work in that sentence. Many questions hand you something that looks quadratic and is not, or something that looks like a mess and turns out to be a perfectly ordinary quadratic. So the first move is always the same: expand every bracket, bring every term to one side, and collect like terms.

Example 1 — Looks quadratic, is not
Is (x − 3)² = x² + 5 a quadratic equation?

Expand the left side: x² − 6x + 9 = x² + 5.
Bring everything to one side: x² − 6x + 9 − x² − 5 = 0, which is −6x + 4 = 0.

The x² terms cancelled, so the highest power is 1. It is a linear equation, not a quadratic one. Never judge by appearance — always simplify first.
Example 2 — Rewriting in standard form
Write x(x + 4) = 3x − 2 in standard form and identify a, b and c.

Expand: x² + 4x = 3x − 2.
Move everything left: x² + 4x − 3x + 2 = 0.
Collect: x² + x + 2 = 0.

So a = 1, b = 1, c = 2.
Example 3 — A cubic that collapses
Is (x + 2)³ = x³ − 4 a quadratic equation?

Expand the left side: x³ + 6x² + 12x + 8 = x³ − 4.
The x³ terms cancel: 6x² + 12x + 12 = 0, or dividing throughout by 6, x² + 2x + 2 = 0.

Yes — it is quadratic. Why it works: the cubic terms were identical on both sides, so they subtracted away. Questions like this are testing whether you simplify before deciding.
Common Mistake — Reading off a, b and c before the equation is in standard form. In 5x − 3 = 2x², the values are not a = 5, b = −3. Rearrange first to 2x² − 5x + 3 = 0, then a = 2, b = −5, c = 3. Signs matter enormously in the formula that follows.

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2. What a Root Really Means

A root (or solution) of a quadratic equation is simply a number that makes the equation true when you put it in place of x. That is the entire definition. If substituting the number gives you 0 = 0, it is a root; if it gives you anything else, it is not.

You met this idea last chapter under a different name. The roots of ax² + bx + c = 0 are exactly the zeroes of the polynomial ax² + bx + c — the places where its graph crosses the x-axis. Same numbers, two vocabularies.

Example 1 — Verifying a root
Check whether x = −2 is a root of x² + 5x + 6 = 0.

Substitute: (−2)² + 5(−2) + 6 = 4 − 10 + 6 = 0.

It gives zero, so yes, x = −2 is a root. Be careful with the squaring of a negative: (−2)² = +4, not −4.
Example 2 — Using a known root to find a constant
If x = 3 is a root of x² − kx + 6 = 0, find k.

Since 3 is a root, substituting it must give zero:
3² − k(3) + 6 = 0
9 − 3k + 6 = 0
15 = 3k, so k = 5.

Check: with k = 5 the equation is x² − 5x + 6 = 0, and 9 − 15 + 6 = 0. ✓
Key Idea — A quadratic equation has at most two roots. It may have two different real roots, one repeated real root, or no real root at all — and Section 5 tells you which, without solving.

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3. Solving by Factorisation (Splitting the Middle Term)

Factorisation is the fastest method when it works, and it rests on one beautifully simple fact: if two numbers multiply to give zero, then at least one of them must be zero. So once you have written the quadratic as (something) × (something) = 0, you can set each bracket to zero separately.

The technique for getting there is called splitting the middle term. Here is the recipe, and it never changes:

  1. Put the equation in standard form ax² + bx + c = 0.
  2. Multiply a × c.
  3. Find two numbers that multiply to a × c and add to b.
  4. Split the middle term into those two pieces.
  5. Group in pairs, take out common factors, and read off the two brackets.
  6. Set each bracket to zero and solve.
Example 1 — The gentlest case (a = 1)
Solve x² + 7x + 12 = 0.

Here a × c = 1 × 12 = 12, and b = 7. Two numbers that multiply to 12 and add to 7: 3 and 4.
Split: x² + 3x + 4x + 12 = 0
Group: x(x + 3) + 4(x + 3) = 0
Factor: (x + 3)(x + 4) = 0
So x + 3 = 0 or x + 4 = 0, giving x = −3 or x = −4.

Check the first one: 9 − 21 + 12 = 0. ✓
Example 2 — When a is not 1
Solve 2x² − 7x + 3 = 0.

a × c = 2 × 3 = 6, and b = −7. Two numbers multiplying to +6 and adding to −7: −6 and −1 (both negative, since their product is positive but their sum is negative).
Split: 2x² − 6x − x + 3 = 0
Group: 2x(x − 3) − 1(x − 3) = 0
Factor: (x − 3)(2x − 1) = 0
x = 3 or x = 1/2.

Why it works: the grouping step only succeeds if both pairs leave the same bracket. If yours do not match, your two numbers were wrong — go back to step 3 rather than pushing on.
Example 3 — Negative constant term
Solve 6x² − 5x − 6 = 0.

a × c = 6 × (−6) = −36, and b = −5. Two numbers multiplying to −36 and adding to −5: −9 and +4.
Split: 6x² − 9x + 4x − 6 = 0
Group: 3x(2x − 3) + 2(2x − 3) = 0
Factor: (2x − 3)(3x + 2) = 0
x = 3/2 or x = −2/3.

Check x = 3/2: 6(9/4) − 5(3/2) − 6 = 13.5 − 7.5 − 6 = 0. ✓
Example 4 — Board level, with a surd coefficient
Solve √2 x² + 7x + 5√2 = 0.

a × c = √2 × 5√2 = 5 × 2 = 10, and b = 7. Two numbers multiplying to 10 and adding to 7: 5 and 2.
Split: √2 x² + 5x + 2x + 5√2 = 0
Group: x(√2 x + 5) + √2(√2 x + 5) = 0   [since √2 × √2 x = 2x and √2 × 5 = 5√2]
Factor: (√2 x + 5)(x + √2) = 0
x = −5/√2 or x = −√2.

Check x = −√2: √2(2) − 7√2 + 5√2 = 2√2 − 7√2 + 5√2 = 0. ✓ Surds look frightening but obey exactly the same rules.
Exam Tip — Always list the factor pairs of a × c systematically before guessing. For 36 that is 1×36, 2×18, 3×12, 4×9, 6×6 — then attach signs. Thirty seconds of listing beats three minutes of trial and error.
Common Mistake — Writing (x − 3)(2x − 1) = 0 and then answering “x = 3 and x = 1”. The second bracket gives 2x = 1, so x = 1/2. Solve each bracket properly rather than reading the number off the page.

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4. The Quadratic Formula and Where It Comes From

Some quadratics simply refuse to factorise with whole numbers. That is not a dead end — it is what the formula is for. It solves every quadratic equation that has real roots, every time, without any cleverness required from you.

Key Rule — For ax² + bx + c = 0 with a ≠ 0, the roots are
x = [ −b ± √(b² − 4ac) ] ÷ 2a, provided b² − 4ac is not negative.

Where does it come from? You rearrange the general equation by completing the square — that is, forcing the left side into a perfect-square shape and then taking the square root of both sides. Your current syllabus asks you to use the formula rather than to reproduce that derivation, but seeing it once makes the formula feel earned instead of memorised.

Using it is mechanical. Write down a, b and c. Compute b² − 4ac on its own line. Then substitute. Doing the discriminant separately, before anything else, prevents most sign errors.

Example 1 — A friendly one, to learn the rhythm
Solve 3x² − 5x + 2 = 0.

a = 3, b = −5, c = 2.
b² − 4ac = (−5)² − 4(3)(2) = 25 − 24 = 1.
x = [5 ± √1] ÷ 6 = (5 ± 1)/6.
So x = 6/6 = 1 or x = 4/6 = 2/3.

x = 1 or x = 2/3. This one also factorises, so you can check: (x − 1)(3x − 2) = 0. ✓
Example 2 — When the answer is irrational
Solve 2x² + x − 4 = 0.

a = 2, b = 1, c = −4.
b² − 4ac = 1 − 4(2)(−4) = 1 + 32 = 33.
x = [−1 ± √33] ÷ 4.

x = (−1 + √33)/4 or x = (−1 − √33)/4.

√33 is not a whole number, and that is completely fine — leave it in surd form. No factorisation with integers could ever have found these, which is exactly why the formula exists.
Example 3 — Equal roots, and a surd coefficient
Solve 4x² + 4√3 x + 3 = 0.

a = 4, b = 4√3, c = 3.
b² − 4ac = (4√3)² − 4(4)(3) = 48 − 48 = 0.
x = [−4√3 ± 0] ÷ 8 = −4√3/8 = −√3/2.

x = −√3/2 (a repeated root). When the square root part is zero, the plus and the minus give the same answer — so the equation has one root, counted twice.

Careful with (4√3)²: square both parts, 16 × 3 = 48.
Common Mistake — Forgetting that the −b at the front applies to the whole coefficient, sign included. In 3x² − 5x + 2 = 0, b is −5, so −b is +5. Writing −5 there flips both roots and loses the question.

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5. The Discriminant and the Nature of Roots

D > 0two distinct rootsD = 0two equal rootsD < 0no real roots
The discriminant D = b² − 4ac decides how many times the parabola meets the x-axis — twice, once, or never.

Sometimes a question does not want the roots at all — it only asks what kind of roots the equation has. The quantity under the square root sign answers that instantly, which is why it gets its own name and symbol.

Key Rule — The discriminant is D = b² − 4ac.
If D > 0: two distinct real roots.
If D = 0: two equal real roots (one repeated root).
If D < 0: no real roots.

The logic is simple once you look at the formula. If D is positive, √D is a real number, and adding and subtracting it gives two different answers. If D is zero, adding and subtracting nothing gives the same answer twice. If D is negative, you are asked for the square root of a negative number, which does not exist among real numbers — so the equation has no real solution.

Example 1 — D positive
Find the nature of the roots of 2x² − 6x + 3 = 0.

D = (−6)² − 4(2)(3) = 36 − 24 = 12.
D > 0, so the equation has two distinct real roots. Since 12 is not a perfect square, those roots are irrational — a nice detail to add if the question says “comment on the roots”.
Example 2 — D zero
Find the nature of the roots of 9x² + 6x + 1 = 0, and find them.

D = 6² − 4(9)(1) = 36 − 36 = 0, so the roots are real and equal.
The repeated root is x = −b/2a = −6/18 = −1/3.

You can see why: 9x² + 6x + 1 is (3x + 1)², a perfect square. Equal roots and perfect squares always travel together.
Example 3 — D negative
Find the nature of the roots of x² + 3x + 5 = 0.

D = 3² − 4(1)(5) = 9 − 20 = −11.
D < 0, so the equation has no real roots.

In your answer write “no real roots” rather than “no solution” — solutions do exist, just not among the real numbers you study this year.
Exam Tip — When D turns out to be a perfect square (1, 4, 9, 16, 25, 36, 49…), the equation would also have factorised. That is a useful signal in the exam: perfect-square D means you can double-check your answer the quick way.

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6. Finding an Unknown Constant (the “Find k” Questions)

These are among the most predictable marks in the whole paper. The question hides a letter inside the coefficients and tells you something about the roots; you convert that statement into a condition on D and solve. There are only three statements they can make.

  • “has equal roots” or “has real and equal roots” → set D = 0.
  • “has real roots” or “has two real roots” → set D ≥ 0.
  • “has no real roots” → set D < 0.
Example 1 — Equal roots
Find k so that kx² − 2√5 x + 4 = 0 has equal roots.

a = k, b = −2√5, c = 4.
D = (−2√5)² − 4(k)(4) = 20 − 16k.
Equal roots means D = 0, so 20 − 16k = 0, giving k = 5/4.

(k = 0 would have destroyed the x² term, so it was never allowed here anyway.)
Example 2 — When the unknown sits in a
Find p so that (p − 3)x² + 6x + 9 = 0 has equal roots.

a = p − 3, b = 6, c = 9.
D = 36 − 4(p − 3)(9) = 36 − 36(p − 3).
Set D = 0: 36 = 36(p − 3), so p − 3 = 1 and p = 4.

One extra line earns you the last mark: p must not equal 3, otherwise the equation would not be quadratic at all. Examiners look for that remark.
Example 3 — An inequality, not an equation
For which values of m does 2x² + mx + 8 = 0 have real roots?

D = m² − 4(2)(8) = m² − 64.
Real roots need D ≥ 0, so m² ≥ 64, which means m ≤ −8 or m ≥ 8.

Why it works: m² ≥ 64 is satisfied by large positive and large negative values of m. Answering only “m ≥ 8” throws away half the answer — a very common slip.

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7. Situational Word Problems

This is where the chapter earns its marks, and where students lose them — almost never because of the algebra, and almost always because of the setting up. Follow the same four steps every single time and word problems become the easiest part of the paper.

  1. Name the unknown. Write “Let x be the smaller number” in words. Never skip this line.
  2. Translate each sentence of the question into algebra, one at a time.
  3. Solve the quadratic by whichever method fits.
  4. Reject impossible roots and answer the question that was actually asked, with units.
Example 1 — A number and its reciprocal
The sum of a number and its reciprocal is 26/5. Find the number.

Let the number be x. Then x + 1/x = 26/5.
Multiply throughout by 5x: 5x² + 5 = 26x, so 5x² − 26x + 5 = 0.
D = 676 − 100 = 576, and √576 = 24.
x = (26 ± 24)/10, giving x = 5 or x = 1/5.

The number is 5 (or 1/5 — both work). Check: 5 + 1/5 = 26/5. ✓ Here both roots are genuine answers, because a number and its reciprocal are interchangeable in the question.
Example 2 — Speed and time
A train travels 300 km at a uniform speed. If its speed had been 5 km/h more, it would have taken 2 hours less for the same journey. Find its speed.

Let the speed be x km/h. Time taken = 300/x hours; at the faster speed, time = 300/(x + 5) hours.
The faster trip is 2 hours shorter: 300/x − 300/(x + 5) = 2.
Multiply through by x(x + 5): 300(x + 5) − 300x = 2x(x + 5)
1500 = 2x² + 10x
x² + 5x − 750 = 0.
D = 25 + 3000 = 3025, and √3025 = 55.
x = (−5 ± 55)/2, giving x = 25 or x = −30.

A speed cannot be negative, so the speed is 25 km/h. Check: 300/25 = 12 hours, 300/30 = 10 hours, a difference of 2. ✓
Example 3 — Area of a rectangle
The length of a rectangular garden is 3 m more than twice its width, and its area is 90 m². Find its dimensions.

Let the width be x metres. Then the length is (2x + 3) metres.
Area: x(2x + 3) = 90, so 2x² + 3x − 90 = 0.
D = 9 + 720 = 729, and √729 = 27.
x = (−3 ± 27)/4, giving x = 6 or x = −15/2.

A width cannot be negative, so the width is 6 m and the length is 2(6) + 3 = 15 m.
Dimensions: 6 m by 15 m. Check: 6 × 15 = 90 m². ✓
Example 4 — An age problem
The product of a girl’s age two years ago and her age four years from now is 27. Find her present age.

Let her present age be x years. Two years ago she was (x − 2); four years from now she will be (x + 4).
(x − 2)(x + 4) = 27
x² + 2x − 8 = 27
x² + 2x − 35 = 0
(x + 7)(x − 5) = 0, so x = −7 or x = 5.

An age cannot be negative, so she is 5 years old. Check: 3 × 9 = 27. ✓
Example 5 — Work and Time (Taps and Cisterns)
Two water taps together can fill a tank in 9⅜ hours. The tap with the larger diameter takes 10 hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank.

Let the smaller tap take x hours to fill the tank. Then the larger tap takes (x − 10) hours.
In 1 hour, they fill 1/x and 1/(x − 10) of the tank. Total time = 9⅜ hours = 75/8 hours.
Equation: 1/x + 1/(x − 10) = 8/75
Multiplying through: 75(x − 10) + 75x = 8x(x − 10)
150x − 750 = 8x² − 80x, so 8x² − 230x + 750 = 0, which simplifies to 4x² − 115x + 375 = 0.
D = 115² − 4(4)(375) = 13225 − 6000 = 7225, and √7225 = 85.
x = (115 ± 85)/8, giving x = 25 or x = 3.75.

If the smaller tap took 3.75 hours, the larger tap would take 3.75 − 10 = −6.25 hours — impossible. So the smaller tap takes 25 hours and the larger tap takes 15 hours. Check: 1/25 + 1/15 = 3/75 + 5/75 = 8/75. ✓
Common Mistake — Giving both roots as the final answer in a real-world problem. Ages, speeds, lengths and numbers of people cannot be negative, and a “smaller part” cannot exceed the whole. Always add one sentence saying which root you rejected and why — CBSE marking schemes award a mark for it.

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8. Which Method Should You Use?

Both methods always give the same roots, so this is a question of speed, not correctness. A simple decision rule serves you well under exam pressure:

  • Compute D = b² − 4ac first. It takes ten seconds.
  • If D is a perfect square, factorisation will work — and it is usually quicker.
  • If D is not a perfect square, go straight to the formula and stop hunting for factors.
  • If the question says “solve by factorisation” or “solve using the quadratic formula”, obey it exactly — marks are attached to the method.
Example 1 — The same equation, both ways
Solve x² − 3x − 10 = 0.

By factorisation: a × c = −10, b = −3, so the numbers are −5 and +2.
x² − 5x + 2x − 10 = 0 → x(x − 5) + 2(x − 5) = 0 → (x − 5)(x + 2) = 0 → x = 5 or x = −2.

By formula: D = 9 + 40 = 49, √49 = 7, so x = (3 ± 7)/2 → x = 5 or x = −2.

Identical, as they must be. Notice D was 49, a perfect square — the signal that factorisation was available.
Example 2 — Always verify at the end
A student solves 2x² − 9x + 4 = 0 and writes x = 4 or x = 1/2. Is the student right?

Substitute x = 4: 2(16) − 9(4) + 4 = 32 − 36 + 4 = 0. ✓
Substitute x = 1/2: 2(1/4) − 9(1/2) + 4 = 0.5 − 4.5 + 4 = 0. ✓

Both check out, so the student is correct. Substituting your answers back takes under a minute and catches nearly every arithmetic slip — make it a habit for the rest of the year.
Good to Know — The CBSE Class 10 Mathematics curriculum for 2026–27 lists this chapter as: standard form ax² + bx + c = 0 (a ≠ 0), solutions of quadratic equations (real roots only) by factorisation and by the quadratic formula, the relationship between the discriminant and the nature of roots, and situational problems from everyday life. Completing the square appears only as the reasoning behind the formula, not as a required method. Do confirm your school’s current syllabus copy if a worksheet asks for something beyond this list.

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Practice Worksheet

Ten questions, arranged from warm-up to board level. Attempt each one fully on paper before you open the answer — the struggle is the part that teaches.

1. Is (x + 2)³ = x³ − 4 a quadratic equation? Justify your answer.
Show Answer
Expand the left side: x³ + 6x² + 12x + 8 = x³ − 4.
The x³ terms cancel, leaving 6x² + 12x + 12 = 0, or x² + 2x + 2 = 0 after dividing by 6.
Yes — it is a quadratic equation, because after simplification the highest power of x is 2 and the coefficient of x² is not zero.
2. Solve by factorisation: 3x² − 14x + 8 = 0.
Show Answer
a × c = 24, b = −14. Numbers: −12 and −2.
3x² − 12x − 2x + 8 = 0
3x(x − 4) − 2(x − 4) = 0
(x − 4)(3x − 2) = 0
x = 4 or x = 2/3.
Check x = 4: 48 − 56 + 8 = 0. ✓
3. Solve using the quadratic formula: 5x² − 6x − 2 = 0.
Show Answer
a = 5, b = −6, c = −2.
D = 36 − 4(5)(−2) = 36 + 40 = 76.
x = [6 ± √76] ÷ 10 = [6 ± 2√19] ÷ 10 = (3 ± √19)/5.
x = (3 + √19)/5 or x = (3 − √19)/5.
(√76 = √(4 × 19) = 2√19 — always simplify the surd.)
4. Find the nature of the roots of 3x² − 4√3 x + 4 = 0, and find the roots if they are real.
Show Answer
D = (−4√3)² − 4(3)(4) = 48 − 48 = 0.
So the roots are real and equal.
The repeated root is x = −b/2a = 4√3/6 = 2/√3 (equivalently 2√3/3).
5. Find the value(s) of k for which x² + kx + 16 = 0 has equal roots.
Show Answer
D = k² − 4(1)(16) = k² − 64.
Equal roots means D = 0, so k² = 64 and k = 8 or k = −8.
Both values are valid answers — giving only k = 8 loses half the marks.
6. The sum of the squares of two consecutive positive integers is 365. Find the integers.
Show Answer
Let the integers be n and n + 1.
n² + (n + 1)² = 365
2n² + 2n + 1 = 365
2n² + 2n − 364 = 0, so n² + n − 182 = 0.
D = 1 + 728 = 729, √729 = 27, so n = (−1 ± 27)/2 = 13 or −14.
The integers are positive, so n = 13 and the numbers are 13 and 14.
Check: 169 + 196 = 365. ✓
7. The length of a rectangular plot exceeds its breadth by 7 m and its area is 144 m². Find its dimensions.
Show Answer
Let the breadth be x metres, so the length is (x + 7) metres.
x(x + 7) = 144 → x² + 7x − 144 = 0.
D = 49 + 576 = 625, √625 = 25, so x = (−7 ± 25)/2 = 9 or −16.
Breadth cannot be negative, so breadth = 9 m and length = 16 m.
Dimensions: 9 m by 16 m. Check: 9 × 16 = 144. ✓
8. Two taps together fill a tank in 6 hours. The larger tap alone takes 5 hours less than the smaller tap alone. How long does each take alone?
Show Answer
Let the smaller tap take x hours alone; the larger takes (x − 5) hours.
In one hour they fill 1/x and 1/(x − 5) of the tank, and together 1/6:
1/x + 1/(x − 5) = 1/6.
Multiply through by 6x(x − 5): 6(x − 5) + 6x = x(x − 5)
12x − 30 = x² − 5x
x² − 17x + 30 = 0
D = 289 − 120 = 169, √169 = 13, so x = (17 ± 13)/2 = 15 or 2.
x = 2 must be rejected (the larger tap would take −3 hours).
Smaller tap: 15 hours; larger tap: 10 hours. Check: 1/15 + 1/10 = 1/6. ✓
9. If one root of 2x² + kx − 6 = 0 is 2, find k and the other root.
Show Answer
Substitute x = 2: 2(4) + 2k − 6 = 0 → 8 + 2k − 6 = 0 → k = −1.
The equation becomes 2x² − x − 6 = 0.
Factorising: 2x² − 4x + 3x − 6 = 0 → 2x(x − 2) + 3(x − 2) = 0 → (x − 2)(2x + 3) = 0.
The other root is x = −3/2.
10. The difference of two numbers is 4 and the difference of their reciprocals is 4/21. Find the numbers.
Show Answer
Let the larger number be x, so the smaller is (x − 4).
The smaller number has the larger reciprocal, so 1/(x − 4) − 1/x = 4/21.
Combine the left side: [x − (x − 4)] ÷ [x(x − 4)] = 4/21, which is 4/(x² − 4x) = 4/21.
So x² − 4x = 21, giving x² − 4x − 21 = 0 → (x − 7)(x + 3) = 0 → x = 7 or x = −3.
Taking the positive solution, the numbers are 7 and 3.
Check: 1/3 − 1/7 = 7/21 − 3/21 = 4/21. ✓

Whatever you scored today, the target for tomorrow is modest and completely achievable: one more correct question than yesterday. Redo the ones that beat you from a blank page, not by rereading the solution. That small daily gain is what turns this chapter into free marks by December.

📝 Practice this chapter: Quadratic Equations — Chapter Test with Answers

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