CBSE Class 10 Mathematics Chapter 4 - Quadratic Equations - Exercise 4.1

 CBSE Class 10 Mathematics  Chapter 4 - Quadratic Equations - Exercise 4.1

Q1.	Check whether the following are quadratic equations :
(i)	(x + 1)2 = 2(x – 3)
Solution	(x + 1)2 = 2(x – 3) By using (a+b)2 = a2 + b2 + 2ab x2 + 12 + 2 (x)(1) = 2(x  – 3) x2 + 1 + 2x = 2x  – 6 x2 + 1 + 2x – 2x + 6 = 0 x2 + 7 = 0  or x2 + 0x + 7 = 0 Form of this equation like standard form of a quadratic equation where a = 1, b = 0 and c = 7 So it is a quadratic equation.

(ii)	x2 – 2x = (–2) (3 – x)
Solution.	x2 – 2x = (–2) (3 – x) x2 – 2x = (–2)3 – (–2)x x2 – 2x = –6 + 2x x2 – 2x + 6 – 2x = 0 x2 – 4x + 6 = 0 Form of this equation like standard form of a quadratic equation where a = 1, b = – 4 and c = 6 So it is a quadratic equation.

(iii)	(x – 2)(x + 1) = (x – 1)(x + 3)
Solution.	(x – 2)(x + 1) = (x – 1)(x + 3) x(x + 1) –2 (x + 1) = x(x + 3) – 1(x + 3) x2 + x – 2x – 2 = x2 + 3x – x – 3 x2 + x – 2x – 2 – x2 – 3x + x + 3 = 0 (x2 – x2) + (x – 2x – 3x + x) – 2 + 3 = 0 0 + (x – 2x – 3x + x) – 2 + 3 = 0 0 + (– 3x) – 2 + 3 = 0 – 3x – 2 + 3 = 0 – 3x + 1 = 0 This equation have highest power is 1 not 2, so it is not a quadratic equation.

(iv)	(x – 3)(2x +1) = x(x + 5)
Solution.	(x – 3)(2x +1) = x(x + 5) x(2x +1) – 3(2x +1) = x(x + 5) 2x2 + x – 6x – 3 = x2 + 5x 2x2 + x – 6x – 3 – x2 – 5x = 0 2x2 – x2 + x – 6x  – 5x – 3 = 0 x2 – 10x – 3 = 0 Form of this equation like standard form of a quadratic equation where a = 1, b = – 10 and c = – 3 So it is a quadratic equation.

(v)	(2x – 1)(x – 3) = (x + 5)(x – 1)
Solution.	(2x – 1)(x – 3) = (x + 5)(x – 1) 2x(x – 3) – 1 (x – 3) = x (x – 1) + 5 (x – 1) 2x2 – 6x – x + 3 = x2 – x + 5x – 5 2x2 – 6x – x + 3 – x2 + x – 5x + 5 = 0 2x2 – x2 – 6x – x + x – 5x + 3 + 5 = 0 x2 – 11x + 8 = 0 Form of this equation like standard form of a quadratic equation where a = 1, b = – 11 and c = 8 So it is a quadratic equation.

(vi)	x2 + 3x + 1 = (x – 2)2
Solution.	x2 + 3x + 1 = (x – 2)2 By using (a – b)2 = a2 + b2 – 2ab x2 + 3x + 1 = x2 + 22 – 2(x)(2) x2 + 3x + 1 = x2 + 4 – 4x x2 + 3x + 1 – x2 – 4 + 4x = 0 x2 – x2 + 3x + 4x + 1 – 4 = 0 7x – 3 = 0 This equation have highest power is 1 not 2, so it is not a quadratic equation.

(vii)	(x + 2)3 = 2x (x2 – 1)
Solution.	(x + 2)3 = 2x (x2 – 1) By using (a + b)3 = a3 + b3 + 3ab (a + b) x3 + 23 + 3(x)(2) (x + 2) = 2x (x2 – 1) x3 + 8 + 6x (x + 2) = 2x3 – 2x x3 + 8 + 6x2 + 12x = 2x3 – 2x x3 + 8 + 6x2 + 12x – 2x3 + 2x = 0 x3 – 2x3 + 6x2 + 12x  + 2x + 8 = 0 – x3 + 6x2 + 14x + 8 = 0 This equation have highest power is 3 not 2, so it is not a quadratic equation.

(viii)

x3 – 4x2 – x + 1 = (x – 2)3

Solution.

x3 – 4x2 – x + 1 = (x – 2)3 By using (a – b)3 = a3 – b3 – 3ab (a – b) x3 – 4x2 – x + 1 = x3 – 23 – 3(x)(2) (x – 2) x3 – 4x2 – x + 1 = x3 – 8 – 6x (x – 2) x3 – 4x2 – x + 1 = x3 – 8 – 6x2 + 12x x3 – 4x2 – x + 1 – x3 + 8 + 6x2 – 12x = 0 x3 – x3 – 4x2 + 6x2 – x – 12x + 1 + 8 = 0 0 + 2x2 – 13x + 8 = 0 2x2 – 13x + 8 = 0 Form of this equation like standard form of a quadratic equation where a = 2, b = – 13 and c = 8 So it is 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 (in metres) is one more than twice its breadth. We need to find the length and breadth of the plot.
Solution.	We know, Area of a rectangle = length × breadth According to question, Area of plot is 528 m2 And length is one meter more than twice its breadth Let breadth of plot = x meters Then length is = (2x + 1) meters So area of plot = length × breadth 528 = (2x + 1) x 528 = 2x2 + x 2x2 + x — 528 = 0 Form of this equation like standard form of a quadratic equation where a = 2, b = 1 and c = —528

(ii)

The product of two consecutive positive integers is 306. We need to find the integers.

Solution.

Let's the first integer as x and the second integer as x + 1( because they are consecutive) As per question their product is 306, so we can write as: First integer × Second integer = Product of integer x(x + 1) = 306 x2 + x = 306 x2 + x — 306 = 0 So it is a quadratic equation where a = 1, b = 1 and c = – 306. For Find out the integers, Split middle term x2 + 18x – 17x – 306 = 0     (because 18 × (–17) = –306 and 18 + (–17) = 1) x(x + 18) – 17(x + 18) = 0 (x – 17)(x + 18) = 0 Now, Set each factor equal to zero and solve x: x – 17 = 0 x = 17 x + 18 = 0 x = – 18 Because we are looking positive integer, so solution is 17.

(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.

Solution.

Let's Rohan’s age = x Rohan’s mother age = x + 26 (Because mother is 26 years older than him) After 3 year their age Rohan age = x + 3 Mother age = (x + 26) + 3 = x + 29 As per question product of their age after 3 year = 360, so we can write as: (x + 3)(x + 29) = 360 Multiply LHS x(x + 29) + 3(x + 29) = 360 x2 + 29x + 3x + 87 = 360 x2 + 29x + 3x + 87 – 360 = 0 x2 + 32x – 273 = 0 So it is a quadratic equation where a = 1, b = 32 and c = – 273. For Find out the Rohan’s present age, Split middle term x2 + 39x – 7x – 273 = 0 x(x + 39) – 7(x + 39) = 0 (x – 7)(x + 39) = 0 Now, Set each factor equal to zero and solve x: x – 7 = 0 x = 7 x + 39 = 0 x = – 39 Because age cannot be negative, so Rohan’s present age is 7 year and his mother present age is 7 + 26 = 33

(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.
Solution.