ICSE Class 10 Maths: Solving Quadratic Equations by Factorisation

Why Factorising Quadratics Is Handy

Ever wondered why a thrown ball follows a perfect arch? That curve hides a quadratic equation you can crack with simple factorisation. Once you master it, those "ax^2+bx+c=0" puzzles become as easy as splitting a LEGO set into two smaller builds.

In simple words: Turn the quadratic into a product of two simple brackets, then tell each bracket to equal zero. The answers you get are the values of x that make the original equation true.

Step‑by‑Step Method to Factorise a Quadratic

Here’s the exact recipe most students use when the exam asks for "solve by factorisation".

• 1. Bring everything to one side. Rewrite the equation so that the right‑hand side is 0. Think of it like moving all the toys to one basket before you start sorting.
• 2. Look for a common factor. If every term shares a number or a variable, pull it out. It’s the same as taking the biggest Lego piece out before you build smaller ones.
• 3. Factor the quadratic. Find two numbers that multiply to give the product of the leading coefficient (the number in front of x²) and the constant term, and that add up to the middle coefficient (the number in front of x). Then split the middle term and group.
• 4. Set each bracket to zero. This uses the zero‑product rule – if a product of two things is zero, at least one of them must be zero.
• 5. Solve the resulting linear equations. Each bracket now looks like "ax + b = 0", which is a straight‑line equation you can solve in a flash.
• 6. Check your answers. Plug the values back into the original equation to make sure they really work.

Worked Example 1: Simple Coefficients

Solve 2x² + 5x – 3 = 0 by factorisation.

1. Equation is already =0.
2. There is no common factor, so move on.
3. We need two numbers that multiply to 2 × (‑3) = ‑6 and add to 5. Those numbers are 6 and ‑1.
4. Rewrite the middle term: 2x² + 6x – x – 3 = 0.
5. Group: (2x² + 6x) + (‑x – 3) = 0 → 2x(x + 3) – 1(x + 3) = 0.
6. Factor out the common bracket (x + 3): (2x – 1)(x + 3) = 0.
7. Apply zero‑product rule: 2x – 1 = 0 OR x + 3 = 0.
8. Solve: x = ½ OR x = ‑3.
9. Check quickly: Both satisfy the original equation, so the solutions are x = ½, ‑3.

Worked Example 2: When a Common Factor Helps

Solve 6x² – 9x = 0.

1. Equation is already =0.
2. Every term shares a factor of 3x. Pull it out: 3x(2x – 3) = 0.
3. Zero‑product rule gives 3x = 0 OR 2x – 3 = 0.
4. Solutions: x = 0 OR x = 3/2.

Quick Summary Table

Common Pitfalls & How to Avoid Them

• Forgetting the sign. When the product a·c is negative, one of the two numbers you look for will be negative. Treat it like balancing a seesaw – one side goes up, the other goes down.
• Mixing up the middle term. Always rewrite the original b x term as the sum of the two numbers you found. It’s like splitting a pizza into two slices that together equal the whole.
• Skipping the check. A quick substitution catches sign errors before the exam.

📝 Likely Exam Questions

1. Solve x² – 7x + 12 = 0 by factorisation.
   Answer: (x‑3)(x‑4)=0 → x=3 or x=4.
2. Find the roots of 4y² – 4y – 3 = 0 using factorisation.
   Answer: (2y+1)(2y‑3)=0 → y=‑½ or y=3/2.
3. Factorise and solve 9t² – 30t + 25 = 0.
   Answer: (3t‑5)²=0 → t=5/3 (double root).
4. Given that one root of 2z² + kz – 15 = 0 is 3, find k and solve the equation.
   Answer: Substitute z=3 → 2·9 + 3k –15 =0 → 18 +3k –15=0 → 3k=‑3 → k=‑1. Equation becomes 2z² – z –15=0 → (2z+5)(z‑3)=0 → z=3 or z=‑5/2.
5. Solve 5x³ – 20x = 0 by first factorising out the common term and then using the zero‑product rule.
   Answer: 5x(x²‑4)=0 → 5x(x‑2)(x+2)=0 → x=0, 2, ‑2.