NCERT Solutions for Class 10 Maths Chapter 4 Exercise 4.2

Understanding Roots and Factorization

A quadratic equation is expressed in the standard form ax² + bx + c = 0, where 'a' is not equal to zero. Solving this equation by factorization relies on the Zero Product Property. This mathematical principle states that if the product of two real expressions is zero (i.e., p × q = 0), then either p = 0, q = 0, or both are zero. In factorization, our primary objective is to rewrite the quadratic polynomial ax² + bx + c as a product of two linear factors, such as (px + q)(rx + s) = 0. Once we split the polynomial, we set each linear factor to zero to obtain the two roots of the quadratic equation. Graphically, these roots represent the points where the parabola intersects the x-axis. Developing fluency in this method requires a strong grasp of integer arithmetic, particularly in identifying two numbers whose product equals the product of 'a' and 'c' (ac), and whose sum equals the middle term coefficient 'b'.

The Splitting the Middle Term Algorithm

1. Identify Coefficients — Write down the given quadratic equation in its standard form: ax² + bx + c = 0. Extract the numerical values of the coefficients a, b, and c.
2. Find Target Product and Sum — Calculate the target product (a × c) and identify the target sum (b). Keep track of their positive or negative signs.
3. Find Factor Pairs — Find two integers, let's call them p and q, such that their product is equal to (a × c) and their sum is equal to b.
4. Split the Middle Term — Replace the middle term bx with (px + qx) in the original equation, expanding it into four terms.
5. Factor by Grouping — Group the terms into two pairs. Extract the greatest common factor (GCF) from each pair to reveal a common binomial factor, then set each linear factor to zero to find the roots.

Avoid Sign Errors in Factorization

The most frequent mistake students make in class 10 maths quadratic equations ex 4 2 is choosing the wrong signs for split factors.

• If ac is positive, both p and q must have the same sign (either both positive or both negative). Look at the sign of b to decide: if b is positive, both factors are positive; if b is negative, both factors are negative.
• If ac is negative, the factors p and q must have opposite signs. The larger absolute factor must carry the same sign as b.

Always perform a quick mental multiplication of your factored binomials to ensure they expand back to the original quadratic expression before solving for x!

Practice Questions with Solutions

• Q: Find the roots of the quadratic equation x² - 3x - 10 = 0 by factorization. A: Step 1: Write down the equation: x² - 3x - 10 = 0. Here, a = 1, b = -3, and c = -10. Step 2: Find two numbers whose product is a × c = 1 × (-10) = -10 and whose sum is b = -3. These numbers are -5 and +2. Step 3: Split the middle term: x² - 5x + 2x - 10 = 0. Step 4: Factor by grouping the first two terms and the last two terms: x(x - 5) + 2(x - 5) = 0. Step 5: Factor out the common binomial (x - 5): (x - 5)(x + 2) = 0. Step 6: Set each factor to zero: x - 5 = 0 or x + 2 = 0. Final answer: The roots of the quadratic equation are x = 5 and x = -2.
• Q: Find the roots of the quadratic equation √2x² + 7x + 5√2 = 0 by factorization. A: Step 1: Identify coefficients: a = √2, b = 7, c = 5√2. Step 2: Determine target product: a × c = √2 × 5√2 = 5 × 2 = 10. The target sum is b = 7. Step 3: Find two numbers that multiply to 10 and add up to 7. These numbers are 5 and 2. Step 4: Split the middle term: √2x² + 2x + 5x + 5√2 = 0. Step 5: Group and factor (note that 2 can be written as √2 × √2): √2x(x + √2) + 5(x + √2) = 0. Step 6: Factor out the common bracket: (x + √2)(√2x + 5) = 0. Step 7: Solve for x: x + √2 = 0 => x = -√2; and √2x + 5 = 0 => x = -5/√2. Final answer: The roots are x = -√2 and x = -5/√2.
• Q: Find two numbers whose sum is 27 and product is 182. A: Step 1: Let the first number be x. Since their sum is 27, the second number is (27 - x). Step 2: Their product is given as 182. Set up the equation: x(27 - x) = 182. Step 3: Expand and rearrange into standard quadratic form: 27x - x² = 182 => x² - 27x + 182 = 0. Step 4: Find two numbers whose product is 182 and sum is -27. The numbers are -13 and -14. Step 5: Split the middle term: x² - 13x - 14x + 182 = 0. Step 6: Factor by grouping: x(x - 13) - 14(x - 13) = 0 => (x - 13)(x - 14) = 0. Step 7: Solve for x: x = 13 or x = 14. Final answer: The two required numbers are 13 and 14.
• Q: The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, find the other two sides. A: Step 1: Let the base of the right triangle be x cm. The altitude becomes (x - 7) cm. Step 2: Apply Pythagoras Theorem: Base² + Altitude² = Hypotenuse². Step 3: Substitute the expressions: x² + (x - 7)² = 13². Step 4: Expand and simplify: x² + (x² - 14x + 49) = 169 => 2x² - 14x + 49 - 169 = 0 => 2x² - 14x - 120 = 0. Step 5: Divide the entire equation by 2: x² - 7x - 60 = 0. Step 6: Find two numbers whose product is -60 and sum is -7. These are -12 and 5. Step 7: Factor the quadratic equation: x² - 12x + 5x - 60 = 0 => x(x - 12) + 5(x - 12) = 0 => (x - 12)(x + 5) = 0. Step 8: Solve: x = 12 or x = -5. Since side length cannot be negative, we discard x = -5. Hence, base = 12 cm. Step 9: Calculate altitude: 12 - 7 = 5 cm. Final answer: The other two sides are 12 cm (base) and 5 cm (altitude).

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