ICSE Class 10 Maths Chapter 07 Problems on Quadratic Equations

Chapter 7

Problems on Quadratic Equations

Points to Remember

1. Method of solving problems on quadratic equations:

(i) Denote the unknown quantity in the given problem by x.

(ii) Form the quadratic equation with the help of the given condition or statement.

(iii) Then solve for x.

Then x will be the required answer.

Note. Lengths, areas, volume, money etc. is always taken as positive, neglecting the negative values.

Exercise 7

Question 1

Find two numbers whose sum is 40 and product 375.

Solution

Sum of two numbers = 40

Let first number = x

Then, second number = 40 - x

According to the condition,

\[x(40 - x) = 375\]

\[\Rightarrow 40x - x^2 = 375\]

\[\Rightarrow 40x - x^2 - 375 = 0\]

\[\Rightarrow -x^2 + 40x - 375 = 0\]

\[\Rightarrow x^2 - 40x + 375 = 0\]

\[\Rightarrow x^2 - 15x - 25x + 375 = 0\]

Since \(-40 = -15 - 25\) and \(375 = (-15)(-25)\)

\[\Rightarrow x(x - 15) - 25(x - 15) = 0\]

\[\Rightarrow (x - 15)(x - 25) = 0\]

[Zero Product Rule]

Either \(x - 15 = 0\), then \(x = 15\)

or \(x - 25 = 0\), then \(x = 25\)

(i) If first number = 15, then second number = 40 - 15 = 25

(ii) If first number = 25, then second number = 40 - 25 = 15

Hence, two required numbers are 15 and 25 Ans.

Teacher's Note

When dividing something like a pizza or money between two people, you use quadratic equations to find how much each person should get based on their relationship or constraints.

Question 2

The difference between two integers is 4. Their product is 221. Find the numbers.

Solution

Let the required natural numbers be x and (4 - x).

Then, \(x(4 - x) = 221\)

\[\Rightarrow 4x - x^2 = 221\]

\[\Rightarrow x^2 - 4x + 221 = 0\]

\[\Rightarrow x^2 - 17x + 13x + 221 = 0\]

\[\Rightarrow x(x - 17) - 13(x - 17) = 0\]

\[\Rightarrow (x - 17)(x - 13) = 0\]

\[\Rightarrow x = 17 \text{ or } x = 13\]

Hence, the required numbers are 17 and 13.

Teacher's Note

Finding two numbers with a specific difference and product is like solving a puzzle; this same technique helps engineers design structures with exact measurements.

Question 3

The sum of a natural number and its reciprocal is \(\frac{65}{8}\). Find the natural number.

Solution

Let the natural number = x

Then, its reciprocal = \(\frac{1}{x}\)

According to the condition,

\[x + \frac{1}{x} = \frac{65}{8}\]

\[\Rightarrow 8x^2 + 8 = 65x\]

[Multiplying by 8x]

\[\Rightarrow 8x^2 - 65x + 8 = 0\]

Since \(8 \times 8 = 64\), \(-65 = -64 - 1\), and \(64 = (-64)(-1)\)

\[\Rightarrow 8x^2 - 64x - x + 8 = 0\]

\[\Rightarrow 8x(x - 8) - 1(x - 8) = 0\]

\[\Rightarrow (x - 8)(8x - 1) = 0\]

[Zero Product Rule]

Either \(x - 8 = 0\), then \(x = 8\)

or \(8x - 1 = 0\), then \(8x = 1 \Rightarrow x = \frac{1}{8}\)

But, it is not possible as it is not a natural number

\(\therefore\) Required number = 8 Ans.

Teacher's Note

Understanding reciprocals helps in situations like calculating efficiency ratios in factories or determining medication dosages in hospitals.

Question 4

Divide 27 into two parts such that the sum of their reciprocals is \(\frac{3}{20}\).

Solution

Let the two parts be x and (27 - x).

Then, \(\frac{1}{x} + \frac{1}{27 - x} = \frac{3}{20}\)

\[\Leftrightarrow \frac{27 - x + x}{x(27 - x)} = \frac{3}{20}\]

\[\Rightarrow 3x(27 - x) = 27 \times 20\]

\[\Rightarrow x(27 - x) = 9 \times 20\]

\[\Rightarrow x^2 - 27x + 180 = 0\]

\[\Rightarrow x^2 - 15x - 12x + 180 = 0\]

\[\Rightarrow x(x - 15) - 12(x - 15) = 0\]

\[\Rightarrow (x - 15)(x - 12) = 0\]

\[\Rightarrow x = 15 \text{ or } x = 12\]

Hence, the required two parts are 15 and 12.

Teacher's Note

Dividing resources or time into parts with specific reciprocal relationships appears in scheduling and resource allocation problems in real-world management.

Question 5

The sum of two numbers is 12 and the sum of their squares is 74. Find the numbers.

Solution

Sum of two numbers = 12

Let first number = x

Then, second number = 12 - x

According to the condition,

\[(x)^2 + (12 - x)^2 = 74\]

\[\Rightarrow x^2 + 144 - 24x + x^2 = 74\]

\[\Rightarrow 2x^2 - 24x + 144 - 74 = 0\]

\[\Rightarrow 2x^2 - 24x + 70 = 0\]

\[\Rightarrow x^2 - 12x + 35 = 0\]

[Dividing by 2]

\[\Rightarrow x^2 - 7x - 5x + 35 = 0\]

Since \(-12 = -7 - 5\) and \(35 = (-7)(-5)\)

\[\Rightarrow x(x - 7) - 5(x - 7) = 0\]

\[\Rightarrow (x - 7)(x - 5) = 0\]

[Zero Product Rule]

Either \(x - 7 = 0\), then \(x = 7\)

or \(x - 5 = 0\), then \(x = 5\)

(i) If x = 7, then first number = 7 and, second number = 12 - 7 = 5

(ii) If x = 5, then first number = 5 and, second number = 12 - 5 = 7

Hence, required numbers are 5, 7 Ans.

Teacher's Note

Finding numbers with specific sum and sum of squares relates to calculating variance in statistics and optimizing measurements in engineering projects.

Question 6

Find two consecutive natural numbers, the sum of whose squares is 145.

Solution

Let first natural number = x

then, second number = x + 1

According to the condition,

\[(x)^2 + (x + 1)^2 = 145\]

\[\Rightarrow x^2 + x^2 + 2x + 1 = 145\]

\[\Rightarrow 2x^2 + 2x + 1 - 145 = 0\]

\[\Rightarrow 2x^2 + 2x - 144 = 0\]

\[\Rightarrow x^2 + x - 72 = 0\]

[Dividing by 2]

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