Ch 4: Quadratic Equations (Part 1)
Based on the latest CBSE rationalized syllabus. Master Factorization, the Quadratic Formula, and the Nature of Roots.
Exercise 4.1 Solutions
Q1. Check whether the following are quadratic equations:
(i) (x + 1)² = 2(x - 3)
Using (a+b)² = a² + 2ab + b²:
x² + 2x + 1 = 2x - 6
x² + 2x - 2x + 1 + 6 = 0
x² + 7 = 0
It is of the form ax² + bx + c = 0 (where a=1, b=0, c=7). Yes, it is a quadratic equation.
(iii) (x - 2)(x + 1) = (x - 1)(x + 3)
x² + x - 2x - 2 = x² + 3x - x - 3
x² - x - 2 = x² + 2x - 3
-3x + 1 = 0
The highest degree of x is 1. No, it is not a quadratic equation.
Q2. Represent the following situations in the form of quadratic equations:
(i) The area of a rectangular plot is 528 m². The length of the plot is one more than twice its breadth.
Let the breadth be x meters.
Then, length = 2x + 1 meters.
Area = Length × Breadth
x(2x + 1) = 528
2x² + x - 528 = 0
Exercise 4.2 (Factorization Method)
Q1. Find the roots of the following quadratic equations by factorisation:
(i) x² - 3x - 10 = 0
Splitting the middle term (-3x into -5x + 2x):
x² - 5x + 2x - 10 = 0
x(x - 5) + 2(x - 5) = 0
(x - 5)(x + 2) = 0
Equating to zero: x - 5 = 0 ⇒ x = 5, and x + 2 = 0 ⇒ x = -2.
(iii) √2x² + 7x + 5√2 = 0
Product = √2 × 5√2 = 10. We need numbers that multiply to 10 and add to 7 (which are 5 and 2).
√2x² + 2x + 5x + 5√2 = 0
√2x(x + √2) + 5(x + √2) = 0
(x + √2)(√2x + 5) = 0
Roots: x = -√2 and x = -5/√2.
Q3. Find two numbers whose sum is 27 and product is 182.
Solution:
Let the first number be x. Then the second number is 27 - x.
Product = 182
x(27 - x) = 182
27x - x² = 182
x² - 27x + 182 = 0
Splitting middle term (-14x and -13x):
x(x - 14) - 13(x - 14) = 0
(x - 14)(x - 13) = 0
The two numbers are 13 and 14.
Exercise 4.3 (Nature of Roots)
💡 Formula: Discriminant D = b² - 4ac.
Q1. Find the nature of the roots of the following quadratic equations. If the real roots exist, find them:
(i) 2x² - 3x + 5 = 0
Here a=2, b=-3, c=5.
D = b² - 4ac = (-3)² - 4(2)(5) = 9 - 40 = -31.
Since D < 0, No real roots exist.
(ii) 3x² - 4√3x + 4 = 0
Here a=3, b=-4√3, c=4.
D = (-4√3)² - 4(3)(4) = 48 - 48 = 0.
Since D = 0, Real and equal roots exist.
Roots = -b/2a = 4√3 / 6 = 2√3 / 3.
Q2. Find the values of k for each of the following quadratic equations, so that they have two equal roots.
(i) 2x² + kx + 3 = 0
For equal roots, D = 0.
b² - 4ac = 0
k² - 4(2)(3) = 0
k² - 24 = 0 ⇒ k² = 24
k = ±√24 = ±2√6.
Important PYQs
Q. A motor boat whose speed is 18 km/h in still water takes 1 hour more to go 24 km upstream than to return downstream to the same spot. Find the speed of the stream.
Solution:
Let the speed of the stream be x km/h.
Speed upstream = (18 - x) km/h
Speed downstream = (18 + x) km/h
Time upstream = 24 / (18 - x)
Time downstream = 24 / (18 + x)
According to the question:
24 / (18 - x) - 24 / (18 + x) = 1
Taking 24 common and taking LCM:
24 [ (18 + x - 18 + x) / (18² - x²) ] = 1
24 (2x) = 324 - x²
x² + 48x - 324 = 0
Factorizing: x² + 54x - 6x - 324 = 0
x(x + 54) - 6(x + 54) = 0
(x - 6)(x + 54) = 0
Speed cannot be negative, so x = 6.
Speed of the stream is 6 km/h.