NCERT Class 10 Maths: Quadratic Equations Exercise 4.3 Explained

Why New Methods? Understanding Completing the Square and Quadratic Formula

Quadratic equations are mathematical expressions of the form ax² + bx + c = 0, where 'a', 'b', and 'c' are real numbers and a ≠ 0. Finding the 'roots' or 'zeros' of a quadratic equation means finding the values of 'x' that satisfy the equation. While factorisation is an intuitive method, it relies on finding two numbers whose product is ac and sum is b. This isn't always straightforward or even possible with rational numbers.

Exercise 4.3 introduces two general methods that work for all quadratic equations:

1. Completing the Square: This method transforms the quadratic equation into a perfect square trinomial on one side, making it easier to isolate 'x' by taking the square root. It's a foundational technique that helps in understanding the derivation of the quadratic formula.
2. Quadratic Formula: Derived directly from the method of completing the square, the quadratic formula x = [-b ± √(b² - 4ac)] / 2a provides a direct way to find the roots. It's often the quickest method once you're comfortable with it. The expression (b² - 4ac), known as the discriminant, tells us about the nature of the roots (real and distinct, real and equal, or no real roots) even before solving the equation. Understanding these methods ensures you can solve any quadratic equation you encounter in your exams.

Step-by-Step: Solving by Completing the Square

1. Step 1: Normalize the Coefficient of x² — Ensure the coefficient of x² is 1. If it's not, divide the entire equation by the coefficient of x². For ax² + bx + c = 0, divide by 'a' to get x² + (b/a)x + (c/a) = 0.
2. Step 2: Move the Constant Term — Shift the constant term to the right-hand side (RHS) of the equation. So, x² + (b/a)x = -(c/a).
3. Step 3: Complete the Square — Take half of the coefficient of 'x' (which is b/a), square it ((b/2a)²), and add this value to both sides of the equation. This makes the LHS a perfect square trinomial: x² + (b/a)x + (b/2a)² = -(c/a) + (b/2a)².
4. Step 4: Factor the LHS and Simplify RHS — The left side can now be written as a perfect square: (x + b/2a)². Simplify the right-hand side. For example, (x + b/2a)² = (b² - 4ac) / 4a².
5. Step 5: Take the Square Root — Take the square root of both sides, remembering to include both positive and negative roots on the RHS: x + b/2a = ±√((b² - 4ac) / 4a²) = ±√(b² - 4ac) / 2a.
6. Step 6: Solve for x — Isolate 'x' to find the two roots: x = -b/2a ± √(b² - 4ac) / 2a. This simplifies to x = [-b ± √(b² - 4ac)] / 2a, which is the quadratic formula!

The Quadratic Formula: Your Universal Solver

As you saw in the last step of completing the square, the quadratic formula is a direct result of that method. For any quadratic equation ax² + bx + c = 0 (where a ≠ 0), the roots are given by:

x = [-b ± √(b² - 4ac)] / 2a

This formula is incredibly powerful because it works for all quadratic equations, without needing to perform the step-by-step process of completing the square every time. The term D = b² - 4ac is called the discriminant. Its value determines the nature of the roots:

• If D > 0: The equation has two distinct real roots. This means there are two different values of 'x' that satisfy the equation.
• If D = 0: The equation has two equal real roots. This means there is exactly one unique value of 'x' that satisfies the equation (it's a repeated root).
• If D < 0: The equation has no real roots. In this case, the square root of a negative number is involved, leading to complex roots (which you will study in higher classes). For Class 10, if D < 0, you simply state that no real roots exist.

Worked Examples from NCERT Exercise 4.3

• Example 1: Solve 2x² - 7x + 3 = 0 by completing the square. Step 1: Divide by the coefficient of x² (which is 2). x² - (7/2)x + (3/2) = 0 Step 2: Move the constant term to the RHS. x² - (7/2)x = -3/2 Step 3: Take half of the coefficient of 'x' (-7/2), square it ((-7/4)²) = 49/16, and add to both sides. x² - (7/2)x + 49/16 = -3/2 + 49/16 Step 4: Factor the LHS and simplify RHS. (x - 7/4)² = -24/16 + 49/16 (x - 7/4)² = 25/16 Step 5: Take the square root of both sides. x - 7/4 = ±√(25/16) x - 7/4 = ±5/4 Step 6: Solve for x. Case 1: x - 7/4 = 5/4 x = 5/4 + 7/4 = 12/4 = 3 Case 2: x - 7/4 = -5/4 x = -5/4 + 7/4 = 2/4 = 1/2 Final Answer: The roots are x = 3 and x = 1/2.
• Example 2: Find the roots of 4x² + 4√3x + 3 = 0 using the quadratic formula. Step 1: Identify a, b, and c. Here, a = 4, b = 4√3, c = 3. Step 2: Calculate the discriminant D = b² - 4ac. D = (4√3)² - 4(4)(3) D = (16 3) - 48 D = 48 - 48 = 0 Step 3: Since D = 0, there are two equal real roots. Apply the quadratic formula x = [-b ± √D] / 2a. x = [-4√3 ± √0] / (2 4) x = -4√3 / 8 Step 4: Simplify the result. x = -√3 / 2 Final Answer: The roots are x = -√3/2 and x = -√3/2 (two equal roots).

YoLearn's Exam Tip: Avoiding Common Mistakes

When solving quadratic equations using the methods in Ex 4.3, students often make a few recurring errors. Be extra careful with:

1. Sign Errors: Pay close attention to negative signs, especially when calculating b² - 4ac or when distributing negative signs after taking the square root. A common mistake is (-b)² becoming -b² instead of b².
2. Fraction Arithmetic: When completing the square, you'll often deal with fractions. Make sure you're confident with adding, subtracting, and squaring fractions. Find a common denominator carefully.
3. Forgetting ±: When taking the square root of both sides, always remember to include both the positive and negative roots (±). This is crucial to find both solutions for 'x'.
4. Discriminant Check: Always calculate the discriminant D = b² - 4ac first. If D < 0, you can immediately conclude there are no real roots, saving you time and preventing errors from trying to find the square root of a negative number.
5. Simplification: After finding the roots, simplify them to their simplest form. For example, 4/8 should be 1/2.

Practice Questions with Solutions

• Q: Find the roots of the quadratic equation 3x² - 5x + 2 = 0 by completing the square. A: Step 1: Divide by 3: x² - (5/3)x + (2/3) = 0. Step 2: Move constant: x² - (5/3)x = -2/3. Step 3: Add (-5/6)² = 25/36 to both sides: x² - (5/3)x + 25/36 = -2/3 + 25/36. Step 4: Factor LHS: (x - 5/6)² = -24/36 + 25/36 = 1/36. Step 5: Take square root: x - 5/6 = ±√(1/36) = ±1/6. Step 6: Solve for x: x = 5/6 ± 1/6. So, x = (5+1)/6 = 6/6 = 1 or x = (5-1)/6 = 4/6 = 2/3. Final answer: The roots are x = 1 and x = 2/3.
• Q: Use the quadratic formula to find the roots of x² + 5x + 5 = 0. A: Step 1: Identify a=1, b=5, c=5. Step 2: Calculate discriminant D = b² - 4ac = 5² - 4(1)(5) = 25 - 20 = 5. Step 3: Since D > 0, there are two distinct real roots. Apply formula: x = [-b ± √D] / 2a = [-5 ± √5] / (2 * 1). Final answer: The roots are x = (-5 + √5) / 2 and x = (-5 - √5) / 2.
• Q: Determine the nature of the roots of the equation 2x² - 3x + 5 = 0. If real roots exist, find them. A: Step 1: Identify a=2, b=-3, c=5. Step 2: Calculate discriminant D = b² - 4ac = (-3)² - 4(2)(5) = 9 - 40 = -31. Step 3: Since D = -31 < 0. Final answer: The equation has no real roots.
• Q: A rectangular park is to be designed whose breadth is 3 m less than its length. Its area is to be 40 m². Find its length and breadth. A: Step 1: Let the length of the park be L meters. Then the breadth is L - 3 meters. Step 2: Area = Length × Breadth. So, L(L - 3) = 40. Step 3: Form the quadratic equation: L² - 3L - 40 = 0. Step 4: Use quadratic formula with a=1, b=-3, c=-40. D = (-3)² - 4(1)(-40) = 9 + 160 = 169. Step 5: L = [-(-3) ± √169] / (2 * 1) = [3 ± 13] / 2. Step 6: L = (3 + 13) / 2 = 16 / 2 = 8 or L = (3 - 13) / 2 = -10 / 2 = -5. Step 7: Length cannot be negative, so L = 8 meters. Step 8: Breadth = L - 3 = 8 - 3 = 5 meters. Final answer: The length of the park is 8 m and the breadth is 5 m.

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