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Chapter 5 · Kerala SSLC Class 10 Maths

Second Degree Equations

Two ways to learn — a crisp formula reference, or a story that makes quadratics feel obvious.

ax²+bx+c=0FactorisationQuadratic formulaDiscriminant
Kerala SSLCClass 10MathsChapter 5 — Second Degree Equations12 min

Real scenarios where quadratics arise naturally — then the maths.

How story mode works

Each scene is a real situation where quadratic equations arise naturally. The maths unfolds from the problem — not the other way around. Tap any dark panel to see the formal formula.

1

The Garden Fence

A farmer has exactly 40 metres of fencing. He wants to enclose a rectangular garden with an area of 75 m². What should the dimensions be?

Let the length be l metres. Since the perimeter is 40 m, we know 2(l + w) = 40, so w = 20 − l.

The area condition gives us: l × w = 75. Substitute w = 20 − l:

l × (20 − l) = 75

20l − l² = 75

l² − 20l + 75 = 0

A quadratic equation appeared naturally — not from a textbook exercise, but from a real constraint. The farmer needs to solve it to know how long each side should be.

2

The Factoring Shortcut

Factorisation is faster than the formula — when it works. The trick is spotting two numbers that multiply to c and add to b.

The pattern

For x² + bx + c = 0, find p and q so that p × q = c and p + q = b. Then the equation factors as (x + p)(x + q) = 0.

Try x² − 7x + 12 = 0. We need two numbers that multiply to +12 and add to −7.

Candidates for +12: 1×12, 2×6, 3×4, (−3)×(−4) ← try negatives

(−3) + (−4) = −7 ✓

So: x² − 7x + 12 = (x − 3)(x − 4) = 0

Setting each factor to zero: x − 3 = 0 gives x = 3, and x − 4 = 0 gives x = 4.

Exam tip: When a = 1, factorisation is almost always faster. Only reach for the formula when the numbers are awkward or a ≠ 1.
3

The Formula as Insurance

Factorisation is elegant — but it does not always work neatly. The quadratic formula is your insurance policy: it works on every quadratic, no matter how messy the numbers.

The challenge

Solve 3x² − 5x + 1 = 0. Try to factorise it — you will find no clean integer pair.

So we use the formula. First, identify a, b, c. Then compute the discriminant.

a = 3, b = −5, c = 1

Δ = b² − 4ac = (−5)² − 4(3)(1) = 25 − 12 = 13

x = (−(−5) ± √13) / (2 × 3)

x = (5 ± √13) / 6

Two answers: x = (5 + √13) / 6 ≈ 1.43, and x = (5 − √13) / 6 ≈ 0.23. Both are irrational — no integer factorisation would have found them.

The order that never fails: (1) Write in standard form. (2) Compute Δ = b² − 4ac. (3) If Δ ≥ 0, apply the formula. If Δ < 0, there are no real roots.

The Big Picture

Every quadratic exam question fits one of three patterns. Master these and Chapter 5 is done.

Q

Solve it — clean numbers

Factorise: (x−p)(x−q) = 0

Q

Solve it — messy numbers

x = (−b ± √(b²−4ac)) / 2a

Q

Nature of roots (don't solve)

Check Δ = b²−4ac only

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