GSEB Class 10 Maths Solutions Chapter 4 Quadratic Equations Exercise 4.1

Question 1. Check whether the following are quadratic equations:
(i) \( (x + 1)^2 = 2(x - 3) \)
(ii) \( x^2 - 2x = (-2)(3 - x) \)
(iii) \( (x - 2)(x + 1) = (x - 1)(x + 3) \)
(iv) \( (x - 3)(2x + 1) = x(x + 5) \)
(v) \( (2x - 1)(x - 3) = (x + 5)(x - 1) \)
(vi) \( x^2 + 3x + 1 = (x - 2)^2 \)
(vii) \( (x + 2)^3 = 2x(x^2 - 1) \)
(viii) \( x^3 - 4x^2 - x + 1 = (x - 2)^3 \)
Answer:
(i) Given equation: \( (x + 1)^2 = 2(x - 3) \)
LHS \( = (x + 1)^2 = x^2 + 2x + 1 \)
RHS \( = 2(x - 3) = 2x - 6 \)
Therefore, \( x^2 + 2x + 1 = 2x - 6 \)
\( x^2 + 2x + 1 - 2x + 6 = 0 \)
\( x^2 + 7 = 0 \)
This equation is in the form \( ax^2 + bx + c = 0 \), where \( a = 1, b = 0, c = 7 \).
Hence, the given equation is a quadratic equation.
(ii) Given equation: \( x^2 - 2x = (-2)(3 - x) \)
LHS \( = x^2 - 2x \)
RHS \( = (-2)(3 - x) = -6 + 2x \)
Therefore, \( x^2 - 2x = -6 + 2x \)
\( x^2 - 2x + 6 - 2x = 0 \)
\( x^2 - 4x + 6 = 0 \)
This equation is in the form \( ax^2 + bx + c = 0 \), where \( a = 1, b = -4, c = 6 \).
Hence, the given equation is a quadratic equation.
(iii) Given equation: \( (x - 2)(x + 1) = (x - 1)(x + 3) \)
LHS \( = (x - 2)(x + 1) = x^2 + x - 2x - 2 = x^2 - x - 2 \)
RHS \( = (x - 1)(x + 3) = x^2 + 3x - x - 3 = x^2 + 2x - 3 \)
Therefore, \( x^2 - x - 2 = x^2 + 2x - 3 \)
\( x^2 - x - 2 - x^2 - 2x + 3 = 0 \)
\( -3x + 1 = 0 \)
This equation is not in the form \( ax^2 + bx + c = 0 \) because the highest power of \( x \) is 1.
Hence, the given equation is not a quadratic equation.
(iv) Given equation: \( (x - 3)(2x + 1) = x(x + 5) \)
LHS \( = (x - 3)(2x + 1) = 2x^2 + x - 6x - 3 = 2x^2 - 5x - 3 \)
RHS \( = x(x + 5) = x^2 + 5x \)
Therefore, \( 2x^2 - 5x - 3 = x^2 + 5x \)
\( 2x^2 - 5x - 3 - x^2 - 5x = 0 \)
\( x^2 - 10x - 3 = 0 \)
This equation is in the form \( ax^2 + bx + c = 0 \), where \( a = 1, b = -10, c = -3 \).
Hence, the given equation is a quadratic equation.
(v) Given equation: \( (2x - 1)(x - 3) = (x + 5)(x - 1) \)
LHS \( = (2x - 1)(x - 3) = 2x^2 - 6x - x + 3 = 2x^2 - 7x + 3 \)
RHS \( = (x + 5)(x - 1) = x^2 - x + 5x - 5 = x^2 + 4x - 5 \)
Therefore, \( 2x^2 - 7x + 3 = x^2 + 4x - 5 \)
\( 2x^2 - 7x + 3 - x^2 - 4x + 5 = 0 \)
\( x^2 - 11x + 8 = 0 \)
This equation is in the form \( ax^2 + bx + c = 0 \), where \( a = 1, b = -11, c = 8 \).
Hence, the given equation is a quadratic equation.
(vi) Given equation: \( x^2 + 3x + 1 = (x - 2)^2 \)
LHS \( = x^2 + 3x + 1 \)
RHS \( = (x - 2)^2 = x^2 - 4x + 4 \)
Therefore, \( x^2 + 3x + 1 = x^2 - 4x + 4 \)
\( x^2 + 3x + 1 - x^2 + 4x - 4 = 0 \)
\( 7x - 3 = 0 \)
This equation is not in the form \( ax^2 + bx + c = 0 \) because the highest power of \( x \) is 1.
Hence, the given equation is not a quadratic equation.
(vii) Given equation: \( (x + 2)^3 = 2x(x^2 - 1) \)
LHS \( = (x + 2)^3 = x^3 + 3(x)^2(2) + 3(x)(2)^2 + (2)^3 = x^3 + 6x^2 + 12x + 8 \)
RHS \( = 2x(x^2 - 1) = 2x^3 - 2x \)
Therefore, \( x^3 + 6x^2 + 12x + 8 = 2x^3 - 2x \)
\( x^3 + 6x^2 + 12x + 8 - 2x^3 + 2x = 0 \)
\( -x^3 + 6x^2 + 14x + 8 = 0 \)
This equation is not in the form \( ax^2 + bx + c = 0 \) because the highest power of \( x \) is 3.
Hence, the given equation is not a quadratic equation.
(viii) Given equation: \( x^3 - 4x^2 - x + 1 = (x - 2)^3 \)
LHS \( = x^3 - 4x^2 - x + 1 \)
RHS \( = (x - 2)^3 = x^3 - (2)^3 - 3(x)^2(2) + 3(x)(2)^2 = x^3 - 8 - 6x^2 + 12x \)
Therefore, \( x^3 - 4x^2 - x + 1 = x^3 - 8 - 6x^2 + 12x \)
\( x^3 - 4x^2 - x + 1 - x^3 + 8 + 6x^2 - 12x = 0 \)
\( 2x^2 - 13x + 9 = 0 \)
This equation is in the form \( ax^2 + bx + c = 0 \), where \( a = 2, b = -13, c = 9 \).
Hence, the given equation is a quadratic equation.
In simple words: To check if an equation is quadratic, simplify both sides and move all terms to one side. If the highest power of 'x' is 2 (like \( ax^2 \)), it's a quadratic equation. If it's 1 (like \( bx \)) or 3 (like \( x^3 \)), it's not.

Exam Tip: Remember to expand all terms carefully and collect like terms. A quadratic equation must always have the highest power of the variable as 2, and no higher powers.

Question 2. Represent the following problems situations in the form of quadratic equations:
(i) The area of a rectangle plot is 528 m². The length of the plot (in metres) is one more than twice its breadth. Find the length and breadth of the plot.
(ii) The product of two consecutive positive integers is 306. Find the integers.
(iii) Rohan's mother is 26 years older than him. The product of their ages 3 years from now will be 360. We would like to find Rohan's present age.
(iv) A train travels a distance of 480 km at uniform speed. If the speed had been 8 km/h less, then it would have taken 3 hours more to cover the same distance. Find the speed of the train.
Answer:
(i) Let the breadth of the rectangular plot be \( x \) meters.
According to the problem, the length of the plot is one more than twice its breadth.
So, the length \( = (2x + 1) \) meters.
The area of a rectangle is given by Length \( \times \) Breadth.
Given area \( = 528 \) m².
Thus, \( (2x + 1)x = 528 \)
\( 2x^2 + x = 528 \)
\( 2x^2 + x - 528 = 0 \)
This is the required quadratic equation. To find the length and breadth:
We solve this equation by the factorization method:
\( 2x^2 + 33x - 32x - 528 = 0 \)
\( x(2x + 33) - 16(2x + 33) = 0 \)
\( (2x + 33)(x - 16) = 0 \)
This means either \( 2x + 33 = 0 \) or \( x - 16 = 0 \).
If \( 2x + 33 = 0 \), then \( x = -\frac{33}{2} \).
If \( x - 16 = 0 \), then \( x = 16 \).
Since breadth cannot be negative, we take \( x = 16 \).
Therefore, the breadth of the rectangle \( = 16 \) m.
And the length \( = 2(16) + 1 = 32 + 1 = 33 \) m.
In simple words: We set up an equation where 'x' is the width and '2x+1' is the length. Their product (area) is 528. Solving this quadratic equation gives us the width as 16 meters and the length as 33 meters.
(ii) Let the two consecutive positive integers be \( x \) and \( (x + 1) \).
According to the problem, their product is 306.
So, \( x(x + 1) = 306 \)
\( x^2 + x = 306 \)
\( x^2 + x - 306 = 0 \)
This is the required quadratic equation. To find the integers:
We solve this equation by the factorization method:
\( x^2 + 18x - 17x - 306 = 0 \)
\( x(x + 18) - 17(x + 18) = 0 \)
\( (x + 18)(x - 17) = 0 \)
This means either \( x + 18 = 0 \) or \( x - 17 = 0 \).
If \( x + 18 = 0 \), then \( x = -18 \).
If \( x - 17 = 0 \), then \( x = 17 \).
Since we need positive integers, we choose \( x = 17 \).
Therefore, the two consecutive positive integers are 17 and \( (17 + 1) = 18 \).
In simple words: If 'x' is the first integer, the next one is 'x+1'. Their multiplication equals 306. We solve the equation to find 'x'. Since numbers must be positive, 'x' is 17, and the next integer is 18.
(iii) Let Rohan's present age be \( x \) years.
Rohan's mother's present age is 26 years older than him, so her age is \( (x + 26) \) years.
After 3 years:
Rohan's age will be \( (x + 3) \) years.
His mother's age will be \( (x + 26 + 3) = (x + 29) \) years.
According to the problem, the product of their ages after 3 years will be 360.
So, \( (x + 3)(x + 29) = 360 \)
\( x^2 + 29x + 3x + 87 = 360 \)
\( x^2 + 32x + 87 = 360 \)
\( x^2 + 32x + 87 - 360 = 0 \)
\( x^2 + 32x - 273 = 0 \)
This is the required quadratic equation. To find Rohan's present age:
We solve this equation by the factorization method:
\( x^2 + 39x - 7x - 273 = 0 \)
\( x(x + 39) - 7(x + 39) = 0 \)
\( (x + 39)(x - 7) = 0 \)
This means either \( x + 39 = 0 \) or \( x - 7 = 0 \).
If \( x + 39 = 0 \), then \( x = -39 \).
If \( x - 7 = 0 \), then \( x = 7 \).
Since age cannot be negative, we choose \( x = 7 \).
Therefore, Rohan's present age is 7 years.
In simple words: We let Rohan's age be 'x'. His mother's age is 'x+26'. In three years, their ages will be 'x+3' and 'x+29'. Multiplying these gives 360. We solve the equation, and because age cannot be negative, Rohan's age is 7 years.
(iv) Let the uniform speed of the train be \( x \) km/h.
The distance covered is 480 km.
Time taken to cover this distance \( = \frac{\text{Distance}}{\text{Speed}} = \frac{480}{x} \) hours.
If the speed had been 8 km/h less, the new speed would be \( (x - 8) \) km/h.
Time taken with reduced speed \( = \frac{480}{x - 8} \) hours.
According to the problem, the train would have taken 3 hours more with the reduced speed.
So, \( \frac{480}{x - 8} - \frac{480}{x} = 3 \)
To solve this, find a common denominator:
\( 480x - 480(x - 8) = 3x(x - 8) \)
\( 480x - 480x + 3840 = 3x^2 - 24x \)
\( 3840 = 3x^2 - 24x \)
Divide the entire equation by 3:
\( 1280 = x^2 - 8x \)
\( x^2 - 8x - 1280 = 0 \)
This is the required quadratic equation. To find the speed of the train:
We solve this equation by the factorization method:
\( x^2 - 40x + 32x - 1280 = 0 \)
\( x(x - 40) + 32(x - 40) = 0 \)
\( (x - 40)(x + 32) = 0 \)
This means either \( x - 40 = 0 \) or \( x + 32 = 0 \).
If \( x - 40 = 0 \), then \( x = 40 \).
If \( x + 32 = 0 \), then \( x = -32 \).
Since speed cannot be negative, we choose \( x = 40 \).
Therefore, the uniform speed of the train is 40 km/h.
In simple words: We represent the train's speed as 'x'. We know that if the speed drops by 8 km/h, the journey takes 3 hours longer. This difference helps us build the quadratic equation \( x^2 - 8x - 1280 = 0 \). Solving it gives us the speed, which is 40 km/h, because speed cannot be negative.

Exam Tip: For word problems, always define your variables clearly and formulate the equation based on the given conditions. Pay attention to units and ensure your final answer makes sense in the context of the problem (e.g., age or speed cannot be negative).