Quadratic equations are distinguished by their standard form as ax² + bx + c = 0, where ‘a’, ‘b’, and ‘c’ are real numbers and ‘a’ is non-zero. This form shows the unique structure of quadratic equations. It distinguishes them from other polynomial equations. Examples of quadratic equations include 2x² + x – 300 = 0 and 4x – 3x² + 2 = 0. It is an essential topic in the CBSE curriculum for Class 10 Maths.
They help us solve many practical problems and theoretical concepts in mathematics. Equations like 2x² – 3x + 1 = 0 or 1 – x² + 300 = 0 stick to the standard quadratic form when arranged in descending order of their degrees. These equations enhance algebraic skills and form a foundation for understanding more complex mathematical concepts in future classes and engineering.
NCERT Solutions For Class 10 Maths Chapter 4 Quadratic Equations Exercise 4.1
1. Check whether the following are quadratic equations :
(i) (x + 1)² = 2(x – 3)
Step 1: Expand (x + 1)²
(x + 1)² = x² + 2x + 1
Step 2: Expand 2(x – 3)
2(x – 3) = 2x – 6
Step 3: Set up the equation
x² + 2x + 1 = 2x – 6
Step 4: Rearrange the equation to standard quadratic form ax² + bx + c = 0
x² + 2x – 2x + 1 + 6 = 0
x² + 7 = 0
Step 5: Check if the equation is quadratic
The equation x² + 7 = 0 is a quadratic equation.
(ii) x² – 2x = (–2) (3 – x)
Step 1: Expand (–2) (3 – x)
(–2) (3 – x) = -6 + 2x
Step 2: Set up the equation
x² – 2x = -6 + 2x
Step 3: Rearrange the equation to standard quadratic form ax² + bx + c = 0
x² – 2x – 2x + 6 = 0
x² – 4x + 6 = 0
Step 4: Check if the equation is quadratic
The equation x² – 4x + 6 = 0 is a quadratic equation.
(iii) (x – 2)(x + 1) = (x – 1)(x + 3)
Step 1: Expand (x – 2)(x + 1)
(x – 2)(x + 1) = x² – x – 2
Step 2: Expand (x – 1)(x + 3)
(x – 1)(x + 3) = x² + 2x – 3
Step 3: Set up the equation
x² – x – 2 = x² + 2x – 3
Step 4: Rearrange the equation to standard quadratic form ax² + bx + c = 0
x² – x – x² – 2x = -3 + 2
-3x = -1
Step 5: Check if the equation is quadratic
The equation -3x = -1 is not a quadratic equation as it is linear.
(iv) (x – 3)(2x +1) = x(x + 5)
Step 1: Expand (x – 3)(2x +1)
(x – 3)(2x +1) = 2x² – 5x – 3
Step 2: Expand x(x + 5)
x(x + 5) = x² + 5x
Step 3: Set up the equation
2x² – 5x – 3 = x² + 5x
Step 4: Rearrange the equation to standard quadratic form ax² + bx + c = 0
2x² – x² – 5x – 5x – 3 = 0
x² – 10x – 3 = 0
Step 5: Check if the equation is quadratic
The equation x² – 10x – 3 = 0 is a quadratic equation.
(v) (2x – 1)(x – 3) = (x + 5)(x – 1)
Step 1: Expand (2x – 1)(x – 3)
(2x – 1)(x – 3) = 2x² – 7x + 3
Step 2: Expand (x + 5)(x – 1)
(x + 5)(x – 1) = x² + 4x – 5
Step 3: Set up the equation
2x² – 7x + 3 = x² + 4x – 5
Step 4: Rearrange the equation to standard quadratic form ax² + bx + c = 0
2x² – x² – 7x – 4
x + 3 + 5 = 0
x² – 11x + 8 = 0
Step 5: Check if the equation is quadratic
The equation x² – 11x + 8 = 0 is a quadratic equation.
(vi) x² + 3x + 1 = (x – 2)²
Step 1: Expand (x – 2)²
(x – 2)² = x² – 4x + 4
Step 2: Set up the equation
x² + 3x + 1 = x² – 4x + 4
Step 3: Rearrange the equation to standard quadratic form ax² + bx + c = 0
x² + 3x – x² + 4x + 1 – 4 = 0
7x – 3 = 0
Step 4: Check if the equation is quadratic
The equation 7x – 3 = 0 is not a quadratic equation as it is linear.
(vii) (x + 2)³ = 2x (x² – 1)
Step 1: Expand (x + 2)³
(x + 2)³ = x³ + 6x² + 12x + 8
Step 2: Expand 2x (x² – 1)
2x (x² – 1) = 2x³ – 2x
Step 3: Set up the equation
x³ + 6x² + 12x + 8 = 2x³ – 2x
Step 4: Rearrange the equation to standard quadratic form ax² + bx + c = 0
x³ – 2x³ + 6x² + 12x + 2x + 8 = 0
– x³ + 6x² + 14x + 8 = 0
Step 5: Check if the equation is quadratic
The equation – x³ + 6x² + 14x + 8 = 0 is not a quadratic equation as it is cubic.
(viii) x³ – 4x² – x + 1 = (x – 2)³
Step 1: Expand (x – 2)³
(x – 2)³ = x³ – 6x² + 12x – 8
Step 2: Set up the equation
x³ – 4x² – x + 1 = x³ – 6x² + 12x – 8
Step 3: Rearrange the equation to standard quadratic form ax² + bx + c = 0
x³ – x³ – 4x² + 6x² – x – 12x + 1 + 8 = 0
2x² – 13x + 9 = 0
Step 4: Check if the equation is quadratic
The equation 2x² – 13x + 9 = 0 is a quadratic equation.
2. 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 (in metres) is one more than twice its breadth. We need to find the length and breadth of the plot.
Step 1: Let the breadth be x meters. Then, the length is (2x + 1) meters.
Step 2: Write the equation for the area of the rectangle.
Area = length × breadth
528 = x × (2x + 1)
Step 3: Formulate the quadratic equation.
528 = 2x² + x
Step 4: Rearrange the equation to standard form ax² + bx + c = 0.
2x² + x – 528 = 0
(ii) The product of two consecutive positive integers is 306. We need to find the integers.
Step 1: Let the first integer be x. Then, the next consecutive integer is x + 1.
Step 2: Write the equation for the product of these integers.
x(x + 1) = 306
Step 3: Formulate the quadratic equation.
x² + x = 306
Step 4: Rearrange the equation to standard form ax² + bx + c = 0.
x² + x – 306 = 0
(iii) Rohan’s mother is 26 years older than him. The product of their ages (in years) 3 years from now will be 360. We would like to find Rohan’s present age.
Step 1: Let Rohan’s present age be x years. Then, his mother’s present age is (x + 26) years.
Step 2: Write the equation for their ages three years from now.
(x + 3)(x + 26 + 3) = 360
Step 3: Formulate the quadratic equation.
(x + 3)(x + 29) = 360
Step 4: Expand and rearrange to standard form ax² + bx + c = 0.
x² + 32x + 87 = 360
x² + 32x – 273 = 0
(iv) A train travels a distance of 480 km at a uniform speed. If the speed had been 8 km/h less, then it would have taken 3 hours more to cover the same distance. We need to find the speed of the train.
Step 1: Let the original speed of the train be x km/h.
Step 2: Write the equation using the relation between speed, distance, and time.
Time = Distance / Speed
Original time = 480 / x
Increased time = 480 / (x – 8)
Step 3: Formulate the equation based on the condition given.
480 / x + 3 = 480 / (x – 8)
Step 4: Rearrange the equation to form a quadratic equation.
Multiplying both sides by x(x – 8) leads to
480(x – 8) + 3x(x – 8) = 480x
480x – 3840 + 3x² – 24x = 480x
Step 5: Simplify and rearrange to standard form ax² + bx + c = 0.
3x² – 24x – 3840 = 0