NCERT Solutions for Class 10 Maths: Quadratic Equations Ex 4.4

The Discriminant: Your Key to Understanding Roots

Every quadratic equation can be written in the standard form: ax² + bx + c = 0, where a, b, and c are real numbers and 'a' is not zero. We know the quadratic formula to find its roots (solutions) is:

x = [ -b ± √(b² - 4ac) ] / 2a

Look closely at the part inside the square root: b² - 4ac. This expression is called the Discriminant, usually denoted by 'D'. The value of the discriminant discriminates or differentiates between the possible types of roots. Think about it: if the value inside a square root is positive, you get a real number. If it's zero, the square root is zero. But if it's negative, you can't find a real square root! This simple observation is the foundation for determining the nature of roots. The discriminant tells us the story of the roots before we even begin to find them.

Interpreting the Discriminant: The Three Cases

Case 1: D > 0 (Discriminant is positive)

If b² – 4ac > 0, the quadratic equation has two distinct real roots. This is because the '±' in the quadratic formula will give two different values: (-b + √D)/2a and (-b - √D)/2a.

Case 2: D = 0 (Discriminant is zero)

If b² – 4ac = 0, the quadratic equation has two equal real roots (or a single real root). The quadratic formula becomes x = -b/2a, as the √D part vanishes. Both roots are identical.

Case 3: D < 0 (Discriminant is negative)

If b² – 4ac < 0, the quadratic equation has no real roots. Since the square root of a negative number is not a real number, we cannot find any real solutions for x. (You will learn about 'imaginary' or 'complex' roots in higher classes).

Worked Examples: Applying the Discriminant

• Question 1: Find the nature of the roots of the quadratic equation 3x² – 5x + 2 = 0. Step 1: Identify a, b, and c. Compare the equation with the standard form ax² + bx + c = 0. Here, a = 3, b = -5, and c = 2. Step 2: Calculate the discriminant (D). D = b² – 4ac D = (-5)² – 4(3)(2) D = 25 – 24 D = 1 Step 3: Analyze the value of D. Since D = 1, which is greater than 0 (D > 0), the equation has two distinct real roots.
• Question 2: Find the value(s) of k for which the quadratic equation kx(x – 2) + 6 = 0 has two equal roots. Step 1: Convert the equation to standard form. First, expand the given equation: kx² – 2kx + 6 = 0 Step 2: Identify a, b, and c. Comparing with ax² + bx + c = 0: a = k, b = -2k, c = 6 Step 3: Apply the condition for equal roots. For an equation to have two equal roots, the discriminant must be zero (D = 0). D = b² – 4ac = 0 Step 4: Substitute and solve for k. (-2k)² – 4(k)(6) = 0 4k² – 24k = 0 4k(k – 6) = 0 This gives two possible solutions: 4k = 0 or k – 6 = 0. So, k = 0 or k = 6. Step 5: Check for validity. If k = 0, the original equation becomes 0(x)(x-2) + 6 = 0, which simplifies to 6 = 0. This is not a quadratic equation. Therefore, k = 0 is not a valid solution. The only valid solution is k = 6.

Board Exam Alert: Avoid These Common Mistakes

1. Sign Errors: Be extremely careful with negative signs when calculating b² - 4ac. A common mistake is forgetting that (-b)² is always positive. For example, if b = -7, b² is (-7)² = 49, not -49.

2. Incorrect a, b, c: Always ensure your equation is in the standard form ax² + bx + c = 0 before you identify the coefficients. If the equation is 2x = 5 - 3x², you must rearrange it to 3x² + 2x - 5 = 0 first. Here a=3, b=2, c=-5.

3. Forgetting to check 'k': In problems where you find a value for 'k', always check if your solution makes sense. If a value of 'k' makes the a coefficient zero (like in our worked example), it's not a valid solution because the equation is no longer quadratic.

Practice Questions with Solutions

• Q: Find the nature of the roots of the quadratic equation 2x² – 6x + 3 = 0. If real roots exist, find them. A: Step 1: Identify a, b, and c. Here, a = 2, b = -6, c = 3. Step 2: Calculate the discriminant, D = b² – 4ac. D = (-6)² – 4(2)(3) = 36 – 24 = 12. Step 3: Analyze D. Since D = 12 > 0, the equation has two distinct real roots. Step 4: Find the roots using the quadratic formula. x = [ -(-6) ± √12 ] / (2*2) = [ 6 ± 2√3 ] / 4. x = [ 3 ± √3 ] / 2. Final answer: The equation has two distinct real roots: (3 + √3)/2 and (3 - √3)/2.
• Q: Find the value of k for the quadratic equation x² – 2x(1 + 3k) + 7(3 + 2k) = 0 to have equal roots. A: Step 1: The equation is already in a form similar to standard form. Identify a, b, and c. a = 1, b = -2(1 + 3k), c = 7(3 + 2k). Step 2: For equal roots, D = 0. So, b² – 4ac = 0. [-2(1 + 3k)]² - 4(1)[7(3 + 2k)] = 0 4(1 + 3k)² - 28(3 + 2k) = 0 Step 3: Solve for k. Divide by 4 to simplify: (1 + 6k + 9k²) - 7(3 + 2k) = 0 1 + 6k + 9k² - 21 - 14k = 0 9k² - 8k - 20 = 0 Step 4: Solve the quadratic in k using factorization or formula. (9k + 10)(k - 2) = 0. k = 2 or k = -10/9. Final answer: The values of k are 2 and -10/9.
• Q: Is it possible to design a rectangular park of perimeter 80 m and area 400 m²? If so, find its length and breadth. A: Step 1: Let the length be 'l' and breadth be 'b'. Perimeter = 2(l + b) = 80, so l + b = 40. This means l = 40 - b. Step 2: Area = l × b = 400. Substitute l = 40 - b. (40 - b) × b = 400 40b - b² = 400 b² - 40b + 400 = 0 Step 3: Check if this situation is possible by finding the nature of roots. Here, a = 1, b = -40, c = 400. D = b² – 4ac = (-40)² – 4(1)(400) = 1600 – 1600 = 0. Step 4: Since D = 0, real roots exist and are equal. So the situation is possible. The root is b = -b / 2a = -(-40) / 2(1) = 20. If breadth b = 20 m, then length l = 40 - 20 = 20 m. Final answer: Yes, it is possible. It will be a square park with length = 20 m and breadth = 20 m.
• Q: Find the discriminant of the equation 9x² - 6x + 1 = 0 and hence find the nature of its roots. A: Step 1: Compare the equation 9x² - 6x + 1 = 0 with ax² + bx + c = 0. We get a = 9, b = -6, c = 1. Step 2: Calculate the discriminant, D. D = b² – 4ac D = (-6)² – 4(9)(1) D = 36 – 36 = 0. Step 3: Analyze the value of D. Since the discriminant is 0, the equation has two equal real roots. Final answer: The discriminant is 0, and the equation has two equal real roots.

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