Quadratic Equations - Exercise 4.1

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Overview

Formulate and test quadratic equations in standard form $ax^2 + bx + c = 0$ ($a \neq 0$). Practice translating real-world area, consecutive integer, age, and train travel scenarios into algebraic equations.

NCERT Exercise Practice questions with detailed step-by-step solutions

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Check Quadratic Form Equations
Question 1 Standard Form & Degree
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) $x^3 - 4x^2 - x + 1 = (x - 2)^3$
(i) $(x + 1)^2 = 2(x - 3)$
Expanding both sides:
$x^2 + 2x + 1 = 2x - 6$
$x^2 + 2x - 2x + 1 + 6 = 0 \implies x^2 + 7 = 0$
Since it is of the standard form $ax^2 + bx + c = 0$ (where $a = 1 \neq 0, b = 0, c = 7$):
$\therefore$ Yes, it is a quadratic equation.
(ii) $x^2 - 2x = (-2)(3 - x)$
Expanding the right-hand side:
$x^2 - 2x = -6 + 2x$
$x^2 - 2x - 2x + 6 = 0 \implies x^2 - 4x + 6 = 0$
This is of the form $ax^2 + bx + c = 0$ (where $a = 1 \neq 0, b = -4, c = 6$):
$\therefore$ Yes, it is a quadratic equation.
(iii) $(x - 2)(x + 1) = (x - 1)(x + 3)$
Expanding both sides:
$x^2 + x - 2x - 2 = x^2 + 3x - x - 3$
$x^2 - x - 2 = x^2 + 2x - 3$
Subtracting $(x^2 + 2x - 3)$ from both sides:
$-x - 2 - 2x + 3 = 0 \implies -3x + 1 = 0 \implies 3x - 1 = 0$
Here the degree of the polynomial is $1$ (no $x^2$ term, $a = 0$):
$\therefore$ No, it is not a quadratic equation.
(iv) $(x - 3)(2x + 1) = x(x + 5)$
Expanding both sides:
$2x^2 + x - 6x - 3 = x^2 + 5x$
$2x^2 - 5x - 3 = x^2 + 5x$
$2x^2 - x^2 - 5x - 5x - 3 = 0 \implies x^2 - 10x - 3 = 0$
It is of the standard form $ax^2 + bx + c = 0$ (where $a = 1 \neq 0$):
$\therefore$ Yes, it is a quadratic equation.
(v) $(2x - 1)(x - 3) = (x + 5)(x - 1)$
Expanding both sides:
$2x^2 - 6x - x + 3 = x^2 - x + 5x - 5$
$2x^2 - 7x + 3 = x^2 + 4x - 5$
$2x^2 - x^2 - 7x - 4x + 3 + 5 = 0 \implies x^2 - 11x + 8 = 0$
It is of the standard form $ax^2 + bx + c = 0$ (where $a = 1 \neq 0$):
$\therefore$ Yes, it is a quadratic equation.
(vi) $x^2 + 3x + 1 = (x - 2)^2$
Expanding RHS using $(a - b)^2 = a^2 - 2ab + b^2$:
$x^2 + 3x + 1 = x^2 - 4x + 4$
Subtracting $(x^2 - 4x + 4)$ from both sides:
$3x + 4x + 1 - 4 = 0 \implies 7x - 3 = 0$
The degree is $1$ (linear equation):
$\therefore$ No, it is not a quadratic equation.
(vii) $(x + 2)^3 = 2x(x^2 - 1)$
Expanding LHS using $(a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$ and expanding RHS:
$x^3 + 3(x^2)(2) + 3(x)(2^2) + 2^3 = 2x^3 - 2x$
$x^3 + 6x^2 + 12x + 8 = 2x^3 - 2x$
Transposing all terms to the left:
$x^3 - 2x^3 + 6x^2 + 12x + 2x + 8 = 0 \implies -x^3 + 6x^2 + 14x + 8 = 0$
The degree of the equation is $3$ (cubic):
$\therefore$ No, it is not a quadratic equation.
(viii) $x^3 - 4x^2 - x + 1 = (x - 2)^3$
Expanding RHS using $(a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3$:
$(x - 2)^3 = x^3 - 3(x^2)(2) + 3(x)(4) - 8 = x^3 - 6x^2 + 12x - 8$
Equating LHS and RHS:
$x^3 - 4x^2 - x + 1 = x^3 - 6x^2 + 12x - 8$
Cancelling $x^3$ from both sides:
$-4x^2 + 6x^2 - x - 12x + 1 + 8 = 0 \implies 2x^2 - 13x + 9 = 0$
Since it has degree $2$ and is of the form $ax^2 + bx + c = 0$ ($a = 2 \neq 0$):
$\therefore$ Yes, it is a quadratic equation.
Question 2 Real-World Mathematical Formulation
Represent the following situations in the form of quadratic equations:
(i) Rectangular Plot Area

Problem: The area of a rectangular plot is $528\text{ m}^2$. 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: Assign variable and express dimensions
Let the breadth of the rectangular plot be $x\text{ metres}$.
Then the length of the plot $= (2x + 1)\text{ metres}$.
Step 2: Formulate quadratic equation from area
$\text{Area} = \text{Length} \times \text{Breadth}$
$(2x + 1)x = 528 \implies 2x^2 + x = 528$
$2x^2 + x - 528 = 0$
Required Equation: $2x^2 + x - 528 = 0$ (where $x$ is the breadth in metres).
(ii) Product of Consecutive Positive Integers

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

Step 1: Define integers algebraically
Let the two consecutive positive integers be $x$ and $(x + 1)$.
Step 2: Set up the product equation
$x(x + 1) = 306$
$x^2 + x = 306 \implies x^2 + x - 306 = 0$
Required Equation: $x^2 + x - 306 = 0$ (where $x$ is the smaller integer).
(iii) Rohan and Mother's Age Relation

Problem: 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: Formulate present and future ages
Let Rohan's present age be $x\text{ years}$.
Then his mother's present age $= (x + 26)\text{ years}$.
After $3\text{ years}$:
Rohan's age $= (x + 3)\text{ years}$
Mother's age $= (x + 26 + 3) = (x + 29)\text{ years}$
Step 2: Apply the product condition
$(x + 3)(x + 29) = 360$
$x^2 + 29x + 3x + 87 = 360$
$x^2 + 32x + 87 - 360 = 0 \implies x^2 + 32x - 273 = 0$
Required Equation: $x^2 + 32x - 273 = 0$ (where $x$ is Rohan's present age in years).
(iv) Train Speed and Travel Time

Problem: A train travels a distance of $480\text{ km}$ at a uniform speed. If the speed had been $8\text{ km/h}$ less, then it would have taken $3\text{ hours}$ more to cover the same distance. We need to find the speed of the train.

Step 1: Express travel times
Let the uniform speed of the train be $x\text{ km/h}$.
Total distance $= 480\text{ km}$.
Time taken at regular speed: $t_1 = \dfrac{480}{x}\text{ hours}$
When speed is decreased by $8\text{ km/h}$, new speed $= (x - 8)\text{ km/h}$.
Time taken at reduced speed: $t_2 = \dfrac{480}{x - 8}\text{ hours}$
Step 2: Formulate the equation from time difference
Given $t_2 - t_1 = 3$:
$\dfrac{480}{x - 8} - \dfrac{480}{x} = 3$
$480 \left(\dfrac{x - (x - 8)}{x(x - 8)}\right) = 3$
$480 \left(\dfrac{8}{x(x - 8)}\right) = 3$
$\dfrac{3840}{x^2 - 8x} = 3$
Dividing both sides by $3$:
$\dfrac{1280}{x^2 - 8x} = 1 \implies x^2 - 8x = 1280 \implies x^2 - 8x - 1280 = 0$
Required Equation: $x^2 - 8x - 1280 = 0$ (where $x$ is the speed of the train in km/h).
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