Chapter 5 · Kerala SSLC Class 10 Maths
Second Degree Equations
Two ways to learn — a crisp formula reference, or a story that builds completing the square from real area and product problems.
Four scenes that build completing-the-square from real problems.
How this works
Each scene turns a real area or product problem into a completed square. Read the story, then tap "The Maths" to see the formal result.
The Growing Square
A square is enlarged by extending each side by 1 metre. Its new area is 100 square metres. What was the length of a side of the original square?
Picture the new square as four pieces: the original green square (side x), two yellow strips of width 1 m running along two sides, and a tiny 1×1 blue square tucked in the corner.
Total area = x² + x + x + 1
= x² + 2x + 1
This must equal 100
So the question becomes: if x² + 2x + 1 = 100, what is x? Does x² + 2x + 1 look familiar? It is exactly the identity from the Class 8 lesson on square identities: (x + 1)² = x² + 2x + 1.
x² + 2x + 1 = 100
(x + 1)² = 100
x + 1 = 10
x = 9
The original square's side was 9 metres.
Completing the Square
Now a harder version: one side of a rectangle is 2 metres longer than the other, and the area is 224 square metres. What are the lengths of the sides?
x(x + 2) = 224
x² + 2x = 224
This time the +1 isn't already sitting there like in Scene 1. But nothing stops us from adding it ourselves — as long as we add it to both sides of the equation, not just one.
x² + 2x + 1 = 224 + 1 = 225
(x + 1)² = 225
x + 1 = 15
x = 14
The shorter side is 14 metres, the longer side 16 metres. This trick — manufacturing the missing piece of a perfect square — is called completing the square.
What if the longer side is 20 metres longer instead of 2 — same area, 224 square metres?
The setup becomes x² + 20x = 224. The same idea works, but now we need a different number to complete the square: half of 20 is 10, and 10² = 100.
x² + 20x = 224
x² + 20x + 100 = 224 + 100 = 324
(x + 10)² = 324
x + 10 = 18
x = 8
This time the sides are 8 metres and 28 metres.
Subtraction and Fractions
The same idea works for subtraction too. A rectangle is cut from a square by trimming 2 metres off one side; the remaining rectangle has area 99 square metres. What was the side of the original square?
x(x − 2) = 99
x² − 2x = 99
Recall the other identity from Class 8: x² − 2x + 1 = (x − 1)². Adding 1 to both sides completes the square here too.
x² − 2x + 1 = 100
(x − 1)² = 100
x − 1 = 10
x = 11
The original square's side was 11 metres.
The coefficient of x isn't even? Does completing the square still work?
Try this: one leg of a right triangle is 5 cm longer than the other, and the triangle's area is 12 square centimetres.
(1/2) × x × (x + 5) = 12
x² + 5x = 24
Half of 5 is a fraction, 5/2 — but it squares just the same way: (5/2)² = 25/4.
x² + 5x + 25/4 = 24 + 25/4 = 121/4
(x + 5/2)² = 121/4
x + 5/2 = 11/2
x = 3
The legs of the triangle are 3 cm and 8 cm.
Two Roads, Two Answers
A rectangle has perimeter 100 metres and area 525 square metres. What are its sides?
Since the perimeter is 100, length + breadth = 50. Let one side be x; the other is then 50 − x.
x(50 − x) = 525
50x − x² = 525
We can't complete the square directly on 50x − x² — it's easier to flip the sign of the whole equation first.
x² − 50x = −525
x² − 50x + 625 = −525 + 625 = 100
(x − 25)² = 100
x − 25 = 10
x = 35
The sides are 35 metres and 15 metres.
Not every second degree equation comes from a shape. Sometimes it just asks: for what x does an expression equal zero? Take p(x) = x² − 4x + 3. When is p(x) = 0?
x² − 4x + 3 = 0
x² − 4x + 4 = 0 + 1 = 1
(x − 2)² = 1
x − 2 = 1 or x − 2 = −1
x = 3 or x = 1
This time both roots are genuine answers — there is no rectangle side or triangle leg here to rule one out. Compare that to the rectangle problems earlier, where a negative root would have to be rejected because a length can't be negative.
One more example, tying back to arithmetic sequences: how many terms of 99, 97, 95, …, starting from the first, must be added to get 900 as the sum?
The nth term is 101 − 2n, and the sum of the first n terms works out to 100n − n².
100n − n² = 900
n² − 100n = −900
n² − 100n + 2500 = −900 + 2500 = 1600
(n − 50)² = 1600
n − 50 = 40 or n − 50 = −40
n = 90 or n = 10
Both answers are valid — adding the first 10 terms or the first 90 terms of this decreasing sequence gives the same sum, 900.
The Big Picture
Every problem in this chapter reduces to the same move: turn an x² + bx expression into a perfect square, then take both square roots.
Already a perfect square?
x² + 2ax + a² = (x + a)²
Missing the constant piece?
Add (half the x-coefficient)² to both sides
Odd or fractional coefficient?
Still add (b/2)² — fractions work the same way
Square equals k — how many roots?
x + a = +√k or −√k, always two
Which root is the real answer?
Check against the problem — reject what doesn't fit
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