Samadhan Academy - Polynomial Worksheet

Samadhan Academy

✦ Excellence in Mathematical Education ✦

Polynomials & Quadratic Equations

Class X | Chapter: Polynomials | Zeroes & Relationships

📚 Instructions for Students

Read each problem carefully and attempt to solve it on your own first. Use rough paper for calculations. Once you've completed your solution, click the "Show Solution" button to verify your answer. Study the step-by-step solutions to understand the methodology.

Identifying Zeroes from Coordinate Points

A quadratic polynomial p(x) passes through the points:

(-6, 0), (0, -30), (4, -20), and (6, 0)

Find the zeroes of the polynomial p(x).

📊 Graph Visualization

★ Complete Solution

Understanding Zeroes:

The zeroes of a polynomial are the x-values where p(x) = 0.

These are the points where the graph intersects the x-axis (where y = 0).

Identifying from Given Points:

Looking at the given points:

Point (-6, 0): Here y = 0, so x = -6 is a zero
Point (0, -30): Here y ≠ 0, so not a zero
Point (4, -20): Here y ≠ 0, so not a zero
Point (6, 0): Here y = 0, so x = 6 is a zero

✓ Answer:

The zeroes of p(x) are -6 and 6

Finding a Polynomial with Irrational Zeroes

Find a quadratic polynomial having zeroes:

α = -52 and β = 52

★ Complete Solution

Formula:

If α and β are the zeroes, then the quadratic polynomial is:

p(x) = k[x² - (α + β)x + αβ]

Step 1: Find Sum of Zeroes

α + β = -52 + 52 = 0

Step 2: Find Product of Zeroes

αβ = (-52)(52)

= -52

Step 3: Form the Polynomial (taking k = 2)

p(x) = 2[x² - 0·x + (-52)]

= 2[x² - 52]

= 2x² - 5

✓ Answer:

The required polynomial is 2x² - 5

(or any non-zero multiple like 4x² - 10, etc.)

Polynomial from Given Sum and Product

Find a quadratic polynomial whose:

Sum of zeroes = -5
Product of zeroes = 6

★ Complete Solution

Given Information:

Sum of zeroes = α + β = -5

Product of zeroes = αβ = 6

Formula:

For a quadratic polynomial with zeroes α and β:

p(x) = k[x² - (sum of zeroes)x + (product of zeroes)]

Solution (taking k = 1):

p(x) = x² - (-5)x + 6

p(x) = x² + 5x + 6

Verification (Factorization):

x² + 5x + 6 = (x + 2)(x + 3)

Zeroes are -2 and -3

Sum = -2 + (-3) = -5 ✓

Product = (-2)(-3) = 6 ✓

✓ Answer:

The required polynomial is x² + 5x + 6

Finding Zeroes by Factorization

Find the zeroes of the polynomial:

p(x) = 3x² + 11x - 4

★ Complete Solution

Method: Factorization by Splitting the Middle Term

Given: 3x² + 11x - 4 = 0

Step 1: Find two numbers whose:

Product = 3 × (-4) = -12
Sum = 11

The numbers are 12 and -1

(Because: 12 × (-1) = -12 and 12 + (-1) = 11)

Step 2: Split the middle term

3x² + 11x - 4 = 3x² + 12x - x - 4

Step 3: Group and factor

= 3x(x + 4) - 1(x + 4)

= (3x - 1)(x + 4)

Step 4: Find zeroes

Setting p(x) = 0:

(3x - 1)(x + 4) = 0

Either 3x - 1 = 0 or x + 4 = 0

x = 13 or x = -4

✓ Answer:

The zeroes are 13 and -4

Evaluating Reciprocal Sum of Zeroes

If α and β are the zeroes of the polynomial:

p(x) = 2x² + 6x - 6

Find the value of 1α + 1β

★ Complete Solution

Given Polynomial:

p(x) = 2x² + 6x - 6

Comparing with ax² + bx + c:

a = 2, b = 6, c = -6

Step 1: Use relationships between zeroes and coefficients

Sum of zeroes: α + β = -ba = -62 = -3

Product of zeroes: αβ = ca = -62 = -3

Step 2: Simplify the required expression

1α + 1β = β + ααβ = α + βαβ

Step 3: Substitute values

1α + 1β = -3-3 = 1

✓ Answer:

1α + 1β = 1

Constructing Polynomial with Transformed Zeroes

If α and β are zeroes of the polynomial:

x² - 3x + 2

Construct a polynomial whose zeroes are 2α + 1 and 2β + 1

★ Complete Solution

Given Polynomial:

x² - 3x + 2

Step 1: Find relationships for original zeroes

Sum of zeroes: α + β = 3

Product of zeroes: αβ = 2

Step 2: Find sum of new zeroes

(2α + 1) + (2β + 1) = 2α + 2β + 2

= 2(α + β) + 2

= 2(3) + 2 = 8

Step 3: Find product of new zeroes

(2α + 1)(2β + 1) = 4αβ + 2α + 2β + 1

= 4αβ + 2(α + β) + 1

= 4(2) + 2(3) + 1

= 8 + 6 + 1 = 15

Step 4: Construct the required polynomial

Using the formula:

p(x) = x² - (sum of zeroes)x + (product of zeroes)

p(x) = x² - 8x + 15

Verification:

Original zeroes: α = 1, β = 2 (from x² - 3x + 2 = (x-1)(x-2))

New zeroes: 2(1)+1 = 3, 2(2)+1 = 5

Check: x² - 8x + 15 = (x-3)(x-5) ✓

✓ Answer:

The required polynomial is x² - 8x + 15

Polynomial with Given Sum and Product

Find a quadratic polynomial whose sum and product of zeroes are:

Sum of zeroes = 0
Product of zeroes = -9

Also find the zeroes of the polynomial.

★ Complete Solution

Given Information:

Sum of zeroes: α + β = 0

Product of zeroes: αβ = -9

Part 1: Construct the Polynomial

Using the formula:

p(x) = k[x² - (α + β)x + αβ]

Taking k = 1:

p(x) = x² - 0·x + (-9)

p(x) = x² - 9

Part 2: Find the Zeroes

Setting p(x) = 0:

x² - 9 = 0

This is a difference of squares:

(x - 3)(x + 3) = 0

Therefore:

x = 3 or x = -3

Verification:

Sum: 3 + (-3) = 0 ✓

Product: 3 × (-3) = -9 ✓

✓ Answer:

Polynomial: x² - 9

Zeroes: 3 and -3

Samadhan Academy - Mathematics Worksheet

Samadhan Academy

✦ Excellence in Education ✦

Mathematics Problem Worksheet

Class X | Quadratic Equations & Applications

Instructions: Attempt each problem independently first. Click the "Show Solution" button to verify your answer. Diagrams are provided where applicable.

Finding Two Positive Numbers

The difference of the squares of two positive numbers is 180. The square of the smaller number is 8 times the greater number.

Find the two numbers.

★ Solution

Let the greater number be $x$ and the smaller number be $y$ .

Given: $x² - y² = 180$ ... (1)

Also: $y² = 8x$ ... (2)

Substituting (2) in (1):

$x² - 8x = 180$

$x² - 8x - 180 = 0$

$(x - 18)(x + 10) = 0$

$x = 18$ or $x = -10$ (rejected, as x is positive)

From (2): $y² = 8 \times 18 = 144$

$y = 12$

Answer: The two numbers are 18 and 12.

Quadratic Equation Coefficients

Express $x +$ 1x $= 3$ as a quadratic equation in the form $ax² + bx + c = 0$ .

Find the value of $a - b + c$ .

★ Solution

Given: $x +$ 1x $= 3$

Multiplying both sides by $x$ :

$x² + 1 = 3x$

$x² - 3x + 1 = 0$

Comparing with $ax² + bx + c = 0$ :

$a = 1, b = -3, c = 1$

Therefore: $a - b + c = 1 - (-3) + 1 = 1 + 3 + 1 = 5$

Answer: $a - b + c$ = 5

Rectangular Field Dimensions

The diagonal of a rectangular field is 60 m more than the shorter side. If the longer side is 30 m more than the shorter side.

Find the length of the sides.

📐 Diagram: Rectangular Field

★ Solution

Let the shorter side = $x$ m

Then: Longer side = $(x + 30)$ m, Diagonal = $(x + 60)$ m

By Pythagoras theorem:

$x² + (x + 30)² = (x + 60)²$

$x² + x² + 60x + 900 = x² + 120x + 3600$

$x² - 60x - 2700 = 0$

$(x - 90)(x + 30) = 0$

$x = 90$ (rejecting negative value)

Shorter side = 90 m, Longer side = 90 + 30 = 120 m

Answer: Shorter side = 90 m, Longer side = 120 m

Fraction Problem

The denominator of a fraction is 2 more than the numerator. If 2 is added to both the numerator and denominator, the sum of the new and original fractions is 4635.

Find the original fraction.

★ Solution

Let the numerator = $x$

Then denominator = $x + 2$

Original fraction = xx + 2

New fraction = x + 2x + 4

Given: xx + 2 + x + 2x + 4 = 4635

Solving: $35[x(x+4) + (x+2)²] = 46(x+2)(x+4)$

$35[x² + 4x + x² + 4x + 4] = 46[x² + 6x + 8]$

$70x² + 280x + 140 = 46x² + 276x + 368$

$24x² + 4x - 228 = 0$

$6x² + x - 57 = 0$

$(6x + 19)(x - 3) = 0$

$x = 3$ (rejecting negative value)

Answer: The original fraction is 35

Age Problem

Sourav's age is 3 years more than the square of Ravi's age. When Ravi reaches Sourav's current age, Sourav will be 6 years less than 13 times Ravi's current age.

Find their present ages.

★ Solution

Let Ravi's present age = $x$ years

Then Sourav's present age = $x² + 3$ years

Years until Ravi reaches Sourav's current age = $(x² + 3) - x = x² - x + 3$

Sourav's age at that time = $(x² + 3) + (x² - x + 3) = 2x² - x + 6$

Given: $2x² - x + 6 = 13x - 6$

$2x² - 14x + 12 = 0$

$x² - 7x + 6 = 0$

$(x - 6)(x - 1) = 0$

$x = 6$ or $x = 1$

If x = 1, Sourav = 4, but Ravi reaching age 4 in 3 years doesn't fit context.

So x = 6: Ravi = 6 years, Sourav = 36 + 3 = 39 years

Answer: Ravi is 6 years old and Sourav is 39 years old.

Weaving Sweaters (LCM Problem)

Ranjita, Neha, and Salma take 15, 18, and 20 days respectively to weave a sweater. They all start on the same day.

When will they all start a new sweater at the same time, and how many sweaters will they have completed in total?

★ Solution

To find when they all start a new sweater together, we need LCM of 15, 18, and 20.

Prime factorization:

15 = 3 × 5

18 = 2 × 3²

20 = 2² × 5

LCM = 2² × 3² × 5 = 4 × 9 × 5 = $180$ days

Sweaters completed:

Ranjita: 180 ÷ 15 = 12 sweaters

Neha: 180 ÷ 18 = 10 sweaters

Salma: 180 ÷ 20 = 9 sweaters

Total = 12 + 10 + 9 = 31 sweaters

Answer: After 180 days, total sweaters completed = 31

2-Digit Number Problem

A 2-digit number (where tens digit > unit digit) is obtained by:

Multiplying the sum of digits by 7 and adding 3, OR
Multiplying the difference of digits by 19 and subtracting 1

Find the 2-digit number.

★ Solution

Let tens digit = $x$ , units digit = $y$ (where x > y)

The number = $10x + y$

From condition 1: $10x + y = 7(x + y) + 3$

$10x + y = 7x + 7y + 3$

$3x - 6y = 3$

$x - 2y = 1$ ... (1)

From condition 2: $10x + y = 19(x - y) - 1$

$10x + y = 19x - 19y - 1$

$-9x + 20y = -1$

$9x - 20y = 1$ ... (2)

From (1): $x = 1 + 2y$

Substituting in (2): $9(1 + 2y) - 20y = 1$

$9 + 18y - 20y = 1$

$-2y = -8$

$y = 4$

Therefore: $x = 1 + 2(4) = 9$

Answer: The 2-digit number is 94

Ratio of Areas: Square and Circle

If the perimeter of a square is equal to the circumference of a circle.

Find the ratio of their areas.

📐 Diagram: Square and Circle Comparison

★ Solution

Let side of square = $a$ , radius of circle = $r$

Given: Perimeter of square = Circumference of circle

$4a = 2πr$

$a =$ πr2

Area of square = $a² =$ π²r²4

Area of circle = $πr²$

Ratio = Area of SquareArea of Circle = π²r²/4πr² = π4

Answer: Ratio of areas (Square : Circle) = π : 4

Circle and Semicircle Radii

The area of a circle is equal to the perimeter of a semicircular disc of equal radius.

Find the radius.

📐 Diagram: Circle and Semicircular Disc

★ Solution

Let the common radius = $r$

Area of circle = $πr²$

Perimeter of semicircular disc = Half circumference + Diameter

= $πr + 2r = r(π + 2)$

Given: Area of circle = Perimeter of semicircular disc

$πr² = r(π + 2)$

$πr = π + 2$ (dividing by r, assuming r ≠ 0)

$r =$ π + 2π $= 1 +$ 2π

≈ 1 + 0.637 ≈ 1.637 units

Answer: $r =$ π + 2π ≈ 1.64 units

Math Worksheet - Samadhan Academy

Samadhan Academy

Mathematics Excellence Series: Step-by-Step Solutions

1. Solve the Rational Equation

Solve for $x$: ; where $ x \neq -4, 7 $

Step 1: Find a common denominator:
$ \frac{(x - 7) - (x + 4)}{(x + 4)(x - 7)} = \frac{11}{30} $

Step 2: Simplify the numerator:
$ \frac{-11}{x^2 - 3x - 28} = \frac{11}{30} $

Step 3: Cross-multiply and solve the quadratic:
$ x^2 - 3x + 2 = 0 $

Answer: $ x = 1 $ or $ x = 2 $

2. Linear Equations: Infinitely Many Solutions

Find $k$ for: $ kx + y = k^2 $ and $ x + ky = 1 $

Condition: $ \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \implies \frac{k}{1} = \frac{1}{k} = \frac{k^2}{1} $

From $ k^2 = 1 $, we get $ k = \pm 1 $. Verifying with the third ratio, only $ k = 1 $ satisfies all parts.

Answer: $ k = 1 $

3. Geometry: Surface Area of a Cone

Assertion (A): CSA of a cone (r=3.5, l=4) is 44 sq cm.
Reason (R): CSA formula is $ \pi rl $.

Calculation: $ CSA = \frac{22}{7} \times 3.5 \times 4 = 11 \times 4 = 44 $.

Conclusion: Both (A) and (R) are true; (R) is the correct explanation.

4. Y-axis Intersection

Equation: $ 2y - x = 4 $

Set $ x = 0 $: $ 2y - 0 = 4 \implies y = 2 $.

Answer: (C) (0, 2)

5. Graphical Representation

Lines: $ 8x - 4y + 12 = 0 $ and $ 2x - y + 5 = 0 $

Compare ratios: $ \frac{8}{2} = 4 $, $ \frac{-4}{-1} = 4 $, $ \frac{12}{5} = 2.4 $.

Since $ \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} $, the lines never meet.

Answer: (C) Parallel

The admission details for B.Sc. programs

https://drive.google.com/file/d/13tBfLcb5kKsquvwheELhIoR8yzutcY5F/view?usp=sharing

Event Announcement:
Samadhan Academy CBSE Classes Inauguration

**Event Announcement: Samadhan Academy CBSE Classes Inauguration**

**Date:** July 25, 2024

**Time:** 10:00 AM

**Location:** University Gate 02, Chhatarpur, M.P. 471001

Samadhan Academy is pleased to announce the inauguration of its new CBSE classes. The event will take place on July 25, 2024, starting at 10:00 AM, at University Gate 02 in Chhatarpur, Madhya Pradesh. This initiative aims to provide high-quality education and academic support to students in the region.

The event will include an introduction to the academy's curriculum, a tour of the facilities, and opportunities to meet the faculty. Prospective students and their parents are invited to attend and learn more about the academy's offerings.

For more information, please contact Samadhan Academy at University Gate 02, Chhatarpur. Call: 9754624944

We look forward to seeing you there!

**Anoop Wari**

**CEO, Samadhan Academy**

University Gate 02

Chhatarpur, M.P. 471001

Phone: 9754624944

### IES Academy : Pure Mathematics Workshop

---

#### Welcome to the IES Academy 2024!

We are thrilled to announce an exciting new addition to our educational offerings – the **Pure Mathematics Workshop**. This workshop is designed for students, educators, and enthusiasts who are passionate about the elegance and depth of pure mathematics. Below, you'll find all the details about this enriching experience.

---

#### Workshop Overview

**Date:** August 1 - August 5, 2024

**Time:** 9:00 AM - 4:00 PM daily

**Venue:** IES Academy, Main Campus, Room 205

This intensive five-day workshop will cover fundamental and advanced topics in pure mathematics, offering participants the opportunity to deepen their understanding and explore new areas of interest.

---

#### Key Topics Covered

- **Number Theory:** Delve into the properties and relationships of numbers, including prime numbers, divisibility rules, and modular arithmetic.

- **Abstract Algebra:** Explore the structures of algebraic systems such as groups, rings, and fields, and their applications.

- **Real Analysis:** Understand the rigorous underpinnings of calculus, including sequences, series, and continuity.

- **Topology:** Learn about the properties of space that are preserved under continuous transformations.

---

#### Featured Speakers

We are honored to host a panel of esteemed mathematicians and educators who will lead the sessions:

- **Anupam Sir**, Professor of Mathematics, IES Academy,

- **Anoop Sir**, Research Scientist, IES Institute of Mathematical Sciences

---

#### **Who Should Attend?**

- **Students:** High school and college students with a keen interest in mathematics.

- **Educators:** Teachers and professors looking to enhance their curriculum and teaching methods.

- **Math Enthusiasts:** Anyone with a passion for mathematics and a desire to learn more about its theoretical aspects.

---

#### Registration Information

**Early Bird Registration (until July 15, 2024):** 200

**Regular Registration (after July 15, 2024):** 250

How to Register:

Visit our [registration page](#) to sign up. Spaces are limited, so be sure to secure your spot early!

---

#### **Contact Us**

For more information, please contact us at:

- **Email:** workshopsiesacademy@gmail.com

- **Phone:** 9754624944

- **Address:** IES Academy, Bajrang nagar , chhatarpur, M..P., 471001

---

We look forward to welcoming you to this transformative experience. Join us at the IES Academy Pure Mathematics Workshop and immerse yourself in the beauty and power of pure mathematics.

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**Stay Connected with IES Academy:**

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Best regards,

**IES Academy Team**

Excellence in Education and Beyond

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WELCOME LINE

📚 Instructions for Students

Identifying Zeroes from Coordinate Points

★ Complete Solution

Finding a Polynomial with Irrational Zeroes

★ Complete Solution

Polynomial from Given Sum and Product

★ Complete Solution

Finding Zeroes by Factorization

★ Complete Solution

Evaluating Reciprocal Sum of Zeroes

★ Complete Solution

Constructing Polynomial with Transformed Zeroes

★ Complete Solution

Polynomial with Given Sum and Product

★ Complete Solution

Finding Two Positive Numbers

★ Solution

Quadratic Equation Coefficients

★ Solution

Rectangular Field Dimensions

★ Solution

Fraction Problem

★ Solution

Age Problem

★ Solution

Weaving Sweaters (LCM Problem)

★ Solution

2-Digit Number Problem

★ Solution

Ratio of Areas: Square and Circle

★ Solution

Circle and Semicircle Radii

★ Solution

1. Solve the Rational Equation

2. Linear Equations: Infinitely Many Solutions

3. Geometry: Surface Area of a Cone

4. Y-axis Intersection

5. Graphical Representation

The admission details for B.Sc. programs

### IES Academy : Pure Mathematics Workshop

#### **Welcome to the IES Academy 2024!**

#### **Workshop Overview**

#### **Key Topics Covered**

#### **Featured Speakers**

#### **Registration Information**

**How to Register:**

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