Quadratic Equations - Exercise 4.2

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Overview

Practice solving quadratic equations by splitting the middle term (factorization). Master sign rules for products and sums, handling irrational coefficients (such as $\sqrt{2}x^2 + 7x + 5\sqrt{2} = 0$), and real-world geometric and commercial word problems.

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

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Q1: Factorisation Q2: Number Pair Q3: Consecutive Integers Q4: Right Triangle Q5: Pottery Industry
Question 1 Factorisation Method
Find the roots of the following quadratic equations by factorisation:
(i) $x^2 - 3x - 10 = 0$
(ii) $2x^2 + x - 6 = 0$
(iii) $\sqrt{2}x^2 + 7x + 5\sqrt{2} = 0$
(iv) $2x^2 - x + \dfrac{1}{8} = 0$
(v) $100x^2 - 20x + 1 = 0$
(i) $x^2 - 3x - 10 = 0$
Find two numbers whose product is $-10$ and sum is $-3$. The numbers are $-5$ and $2$.
Splitting the middle term:
$x^2 - 5x + 2x - 10 = 0$
$x(x - 5) + 2(x - 5) = 0 \implies (x - 5)(x + 2) = 0$
Either $x - 5 = 0 \implies x = 5$ or $x + 2 = 0 \implies x = -2$.
$\therefore$ The roots are $5$ and $-2$.
(ii) $2x^2 + x - 6 = 0$
Product $= 2 \times (-6) = -12$, sum $= +1$. The numbers are $4$ and $-3$.
$2x^2 + 4x - 3x - 6 = 0$
$2x(x + 2) - 3(x + 2) = 0 \implies (2x - 3)(x + 2) = 0$
Either $2x - 3 = 0 \implies x = \dfrac{3}{2}$ or $x + 2 = 0 \implies x = -2$.
$\therefore$ The roots are $\dfrac{3}{2}$ and $-2$.
(iii) $\sqrt{2}x^2 + 7x + 5\sqrt{2} = 0$
Product $= \sqrt{2} \times 5\sqrt{2} = 5 \times 2 = 10$, sum $= 7$. The numbers are $5$ and $2$.
$\sqrt{2}x^2 + 2x + 5x + 5\sqrt{2} = 0$
$\sqrt{2}x(x + \sqrt{2}) + 5(x + \sqrt{2}) = 0 \implies (\sqrt{2}x + 5)(x + \sqrt{2}) = 0$
Either $\sqrt{2}x + 5 = 0 \implies x = -\dfrac{5}{\sqrt{2}} = -\dfrac{5\sqrt{2}}{2}$ or $x + \sqrt{2} = 0 \implies x = -\sqrt{2}$.
$\therefore$ The roots are $-\dfrac{5}{\sqrt{2}}$ and $-\sqrt{2}$.
(iv) $2x^2 - x + \dfrac{1}{8} = 0$
Multiplying the entire equation by $8$:
$16x^2 - 8x + 1 = 0$
Recognizing this as a perfect square $(4x - 1)^2 = 0$ or splitting $-8x = -4x - 4x$:
$4x(4x - 1) - 1(4x - 1) = 0 \implies (4x - 1)(4x - 1) = 0$
$4x - 1 = 0 \implies x = \dfrac{1}{4}$
$\therefore$ The roots are $\dfrac{1}{4}$ and $\dfrac{1}{4}$ (equal roots).
(v) $100x^2 - 20x + 1 = 0$
Recognizing as a perfect square $(10x - 1)^2 = 0$ or splitting $-20x = -10x - 10x$:
$10x(10x - 1) - 1(10x - 1) = 0 \implies (10x - 1)(10x - 1) = 0$
$10x - 1 = 0 \implies x = \dfrac{1}{10}$
$\therefore$ The roots are $\dfrac{1}{10}$ and $\dfrac{1}{10}$ (equal roots).
Question 2 Number Relations
Find two numbers whose sum is $27$ and product is $182$.
Step 1: Assign variable
Let one of the numbers be $x$.
Since their sum is $27$, the other number is $(27 - x)$.
Step 2: Form and solve the quadratic equation
Given their product is $182$:
$x(27 - x) = 182$
$27x - x^2 = 182 \implies x^2 - 27x + 182 = 0$
Splitting $-27x = -13x - 14x$ (since $(-13) \times (-14) = 182$):
$x(x - 13) - 14(x - 13) = 0 \implies (x - 13)(x - 14) = 0$
$\therefore x = 13$ or $x = 14$.
If one number is $13$, the other is $27 - 13 = 14$.
The required two numbers are 13 and 14.
Question 3 Consecutive Integers
Find two consecutive positive integers, sum of whose squares is $365$.
Step 1: Define consecutive integers
Let the two consecutive positive integers be $x$ and $(x + 1)$.
Step 2: Set up and simplify equation
$x^2 + (x + 1)^2 = 365$
$x^2 + x^2 + 2x + 1 = 365$
$2x^2 + 2x + 1 - 365 = 0 \implies 2x^2 + 2x - 364 = 0$
Dividing the entire equation by $2$:
$x^2 + x - 182 = 0$
Step 3: Factorise and solve
$x^2 + 14x - 13x - 182 = 0$
$x(x + 14) - 13(x + 14) = 0 \implies (x + 14)(x - 13) = 0$
Since $x$ must be a positive integer, $x \neq -14$.
$\therefore x = 13$, and the next integer is $13 + 1 = 14$.
The two consecutive positive integers are 13 and 14.
Question 4 Right Triangle & Pythagoras
The altitude of a right triangle is $7\text{ cm}$ less than its base. If the hypotenuse is $13\text{ cm}$, find the other two sides.
Step 1: Assign variables to sides
Let the base of the right triangle be $x\text{ cm}$.
Then the altitude $= (x - 7)\text{ cm}$.
Hypotenuse $= 13\text{ cm}$.
Step 2: Apply the Pythagoras Theorem
$\text{Base}^2 + \text{Altitude}^2 = \text{Hypotenuse}^2$
$x^2 + (x - 7)^2 = 13^2$
$x^2 + x^2 - 14x + 49 = 169$
$2x^2 - 14x + 49 - 169 = 0 \implies 2x^2 - 14x - 120 = 0$
Dividing by $2$:
$x^2 - 7x - 60 = 0$
Step 3: Factorise and solve for side lengths
$x^2 - 12x + 5x - 60 = 0$
$x(x - 12) + 5(x - 12) = 0 \implies (x - 12)(x + 5) = 0$
Since side lengths must be positive, $x \neq -5$.
$\therefore x = 12\text{ cm}$.
$\text{Base} = 12\text{ cm}$ and $\text{Altitude} = 12 - 7 = 5\text{ cm}$.
Base = 12 cm, Altitude = 5 cm (the other two sides are 12 cm and 5 cm).
Question 5 Production & Commercial Math
A cottage industry produces a certain number of pottery articles in a day. It was observed on a particular day that the cost of production of each article (in rupees) was $3$ more than twice the number of articles produced on that day. If the total cost of production on that day was ₹$90$, find the number of articles produced and the cost of each article.
Step 1: Assign variable and express unit cost
Let the number of articles produced on that day be $x$.
Then the cost of production of each article $= ₹(2x + 3)$.
Step 2: Formulate total cost equation
$\text{Total Cost} = (\text{Number of articles}) \times (\text{Cost per article}) = 90$
$x(2x + 3) = 90$
$2x^2 + 3x - 90 = 0$
Step 3: Solve by factorisation
Product $= 2 \times (-90) = -180$, sum $= +3$. Numbers are $15$ and $-12$.
$2x^2 + 15x - 12x - 90 = 0$
$x(2x + 15) - 6(2x + 15) = 0 \implies (2x + 15)(x - 6) = 0$
Since the number of articles must be a positive integer, $x \neq -\dfrac{15}{2}$.
$\therefore x = 6$.
Cost of each article $= 2(6) + 3 = 12 + 3 = ₹15$.
Number of articles produced = 6, Cost of each article = ₹15.
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