197-NCERT New Syllabus Grade 10 Quadratic Equation Ex-4.3

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NCERT New Syllabus Mathematics

Class: 10

Exercise 4.3

Topic: Quadratic Equation

Understanding Quadratic Equations in the New NCERT Syllabus for Class 10

Quadratic equations form a fundamental concept in algebra, crucial for building a strong mathematical foundation. In the new Class 10 syllabus of NCERT, quadratic equations take center stage, preparing students to solve real-world problems efficiently using algebraic techniques.

A quadratic equation is a polynomial equation of degree two, expressed in the form:

ax² + bx + c = 0 where a, b, and c are real numbers, with a ≠ 0.

In this blog, we will explore the various methods of solving quadratic equations, including factorization, completing the square, and the quadratic formula. We’ll also dive into the geometric significance of these equations and how they apply to practical scenarios.

Let’s start by breaking down the key components of quadratic equations and how to solve them step by step!

EXERCISE 4.3

Q1. Find the nature of the roots of the following quadratic equations. 

If the real roots exist, find them:

(i) 2x² – 3x + 5 = 0 (ii) 3x² – 4√3 x + 4 = 0 (iii) 2x² – 6x + 3 = 0 

Explanation:

1) In a quadratic equation of the form ax² + bx + c = 0, where a ≠ 0, the expression

(b² – 4ac) is known as discriminant.

2) Based on the discriminant, the nature of the roots of the quadratic equation 

ax² + bx + c = 0 can be determined as follows:

(a) If b² – 4ac > 0, the equation has two distinct real roots.

(b) If b² – 4ac = 0, the equation has two equal real roots.

(c) If b² – 4ac < 0, the equation has no real roots.

Solution:

(i) 2x² – 3x + 5 = 0

1) The given equation is 2x² – 3x + 5 = 0 ------------------ equation 1.

2) Equate the coefficient of equation 2x² – 3x + 5 = 0 with ax² + bx + c = 0, 

we have,

a = 2, b = – 3, c = 5.

3) First we will find:

b² – 4ac = (– 3)² – 4(2)(5)

b² – 4ac = 9 – 40

b² – 4ac = – 31 ------------------ equation 2. 

4) Since b² – 4ac < 0, the quadratic equation 2x² – 3x + 5 = 0 has no real roots.

(ii) 3x² – 4√3 x + 4 = 0

1) The given equation is 3x² – 4√3 x + 4 = 0 ------------------ equation 1.

2) Equate the coefficient of equation 3x² – 4√3 x + 4 = 0 with ax² + bx + c = 0, 

we have,

a = 3, b = – 4√3, c = 4.

3) First we will find:

b² – 4ac = (– 4√3)² – 4(3)(4)

b² – 4ac = 48 – 48

b² – 4ac = 0 ------------------ equation 2. 

4) As, b² – 4ac = 0, it has two equal real roots, so from equation 2 and equation 3,

we have,

x = [– b ± √(b² – 4ac)]/2a
x = [– (– 4√3) ± √0]/2(3)
x = (4√3 ± 0)/2(3)

x = 2(√3)/3

x = 2(√3)/(√3√3)

x = 2/√3

x = (2√3)/3 

 5) Since, b² – 4ac = 0, the equation has two equal real roots. 

So, the roots are x = (2√3)/3 or x = (2√3)/3.

(iii) 2x² – 6x + 3 = 0

1) The given equation is 2x² – 6x + 3 = 0 ------------------ equation 1.

2) Equate the coefficient of equation 2x² – 6x + 3 = 0 with ax² + bx + c = 0, 

we have,

a = 2, b = – 6, c = 3.

3) First we will find:

b² – 4ac = (– 6)² – 4(2)(3)

b² – 4ac = 36 – 24

b² – 4ac = 12 ------------------ equation 2. 

4) As, b² – 4ac ≥ 0, it has two distinct real roots, so from equation 2 and equation 3,

we have,

x = [– b ± √(b² – 4ac)]/2a
x = [– (– 6) ± √12]/2(2)
x = (6 ± √12)/4

x = (6 ± 2√3)/4

x = (3 ± √3)/2

5) Since b² – 4ac > 0, the equation has two distinct real roots.

So, the roots are x = (3 + √3)/2 or x = (3 – √3)/2.

Q2. Find the values of k for each of the following quadratic equations, so that

they have two equal roots.

(i) 2x² + kx + 3 = 0 (ii) kx (x – 2) + 6 = 0

Explanation:

1) In a quadratic equation of the form ax² + bx + c = 0, where a ≠ 0, the expression

(b² – 4ac) is known as discriminant.

2) Based on the discriminant, the nature of the roots of the quadratic equation 

ax² + bx + c = 0 can be determined as follows:

(a) If b² – 4ac = 0, the equation has two equal real roots.

Solution:

(i) 2x² + kx + 3 = 0

1) The given equation is 2x² + kx + 3 = 0 ------------------ equation 1.

2) Equate the coefficient of equation 2x² + kx + 3 = 0 with ax² + bx + c = 0, 

we have,

a = 2, b = k, c = 3.

3) First we will find:

b² – 4ac = (k)² – 4(2)(3)

b² – 4ac = k² – 24 ------------------ equation 2. 

4) As the quadratic equation has two equal roots, 

b² – 4ac = 0

k² – 24 = 0

k² = 24 

k = ± √24 = ± 2√6

 5) Therefore, k = 2√6 or k = – 2√6.

(ii) kx (x – 2) + 6 = 0

1) The given equation is 

kx (x – 2) + 6 = 0

kx² – 2kx + 6 = 0 ------------------ equation 1.

2) Equate the coefficient of equation kx² – 2kx + 6 = 0 with ax² + bx + c = 0, 

we have,

a = k, b = – 2k, c = 6.

3) First we will find:

b² – 4ac = (– 2k)² – 4(k)(6)

b² – 4ac = 4k² – 24k ------------------ equation 2. 

4) As the quadratic equation has two equal roots, 

b² – 4ac = 0

4k² – 24k = 0

4k(k – 6) = 0
k(k – 6) = 0

 5) So,  k = 0 or k = 6.

Q3. Is it possible to design a rectangular mango grove whose length is twice

its breadth and the area is 800 m²? If so, find its length and breadth.

1) Let the breadth of a rectangular mango grove be x m.

2) So, the length of a rectangular mango grove will be 2x m 

3) According to the problem, the area is 800, so

2x(x) = 800

2x² = 800

x² = 400

x² – 400 = 0 ------------------ equation 1.

4) Equate the coefficient of equation x² – 400 = 0 with ax² + bx + c = 0, 

we have,

a = 1, b = 0, c = – 400.

5) First we will find:

b² – 4ac = (0)² – 4(1)(– 400)

b² – 4ac = 1600 ------------------ equation 2.

6) As b² – 4ac = 1600 > 0, it has real roots, so from equation 1 and equation 2, 

we have,

x = [– b ± √(b² – 4ac)]/2a
x = [– (0) ± √1600]/2

x = (0 ± 40)/2 

7) So,  x = (40)/2 or x = (– 40)/2, i.e. x = 20, or x = – 20.

8) As length is always positive, ignore x = – 20.

9) So, the breadth of the rectangular mango grove is 20 m and 

its length is 40 m.

Q4. Is the following situation possible? If so, determine their present ages.

The sum of the ages of two friends is 20 years. Four years ago, the product of their age in years was 48.

1) Let the present age of the first friend be x.

2) So, the present age of the second friend will be (20 – x).

3) 4 years ago, their ages will be (x – 4) and (20 – x – 4).

4) According to the problem,

(x – 4)(16 – x) = 48

x(16 – x) – 4(16 – x) = 48

16x – x² – 64 + 4x = 48

20x – x² – 64 – 48 = 0
20x – x² – 112 = 0
x² – 20x + 112 = 0 ------------------ equation 1.

5) Equate the coefficient of equation x² – 20x + 112 = 0 with ax² + bx + c = 0, so

a = 1, b = – 20, c = 112

6) First we will find:

b² – 4ac = (– 20)² – 4(1)(112)

b² – 4ac = 400 – 448

b² – 4ac = – 48 ------------------ equation 2.

7) Since b² – 4ac = – 48 < 0, it has no real roots, so the given situation is

impossible.

Q5. Is it possible to design a rectangular park of perimeter 80 m and an area

of 400 m²? If so, find its length and breadth.

1) Let the breadth of a rectangular park be x m.

2) As, the perimeter of a rectangular park = 80 m.

3) So, length = [(perimeter/2) – breadth]

length = [(80/2) – x]

length = (40 – x)

4) According to the problem,

x(40 – x) = 400

40x – x² = 400

x² – 40x + 400 = 0 ------------------ equation 1.

5) Equate the coefficient of equation x² – 40x + 400 = 0 with ax² + bx + c = 0, so

a = 1, b = – 40, c = 400

6) First we will find:

b² – 4ac = (– 40)² – 4(1)(400)

b² – 4ac = 1600 – 1600

b² – 4ac = 0 ------------------ equation 2.

7) As, b² – 4ac = 0, it has two equal real roots, so from equation 1 and equation 2,

we have,

x = [– b ± √(b² – 4ac)]/2a
x = [– (– 40) ± √0]/2(1)
x = (40 ± 0)/2

x = (20 ± 0)

8) Therefore,  x = 20, the breadth = 20 m and the length = 40 – 20 = 20 m.

Conclusion: Unlocking the Power of Quadratic Equations

As we conclude our exploration of Quadratic Equations, we’ve unlocked a crucial mathematical tool that helps us solve various complex problems, from calculating areas to predicting trajectories. Mastering the factorization method, completing the square, and applying the quadratic formula equips you with versatile techniques to tackle challenges beyond the classroom. Embrace these concepts, as they lay the groundwork for higher-level math and real-world applications. Keep pushing your limits, and remember, every problem has a solution—just like every quadratic equation has its roots!

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#NCERTMaths #QuadraticEquations #Class10Mathematics #NewSyllabus2024 #MathConcepts #PolynomialEquations #EquationSolving #MathematicsLearning #CBSEClass10 #Education #Algebra #MathForStudents #MathTips #MathPractice #MathGenius #NCERTClass10 #AlgebraSkills #RootsAndSolutions #MathExploration #FutureMathematicians #UnlockYourPotential

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NCERT New Syllabus Class 10 - Arithmetic Progressions Exercise 5.1

Anil Satpute

196-NCERT New Syllabus Grade 10 Quadratic Equation Ex-4.2

Click here to explore the previous ⇨ NCERT New Syllabus Class 10 - Quadratic Equation Exercise 4.1

NCERT New Syllabus Mathematics

Class: 10

Exercise 4.2

Topic: Quadratic Equation

Understanding Quadratic Equations in the New NCERT Syllabus for Class 10

Quadratic equations form a fundamental concept in algebra, crucial for building a strong mathematical foundation. In the new Class 10 syllabus of NCERT, quadratic equations take center stage, preparing students to solve real-world problems efficiently using algebraic techniques.

A quadratic equation is a polynomial equation of degree two, expressed in the form:

ax² + bx + c = 0 where a, b, and c are real numbers, with a ≠ 0.

In this blog, we will explore the various methods of solving quadratic equations, including factorization, completing the square, and the quadratic formula. We’ll also dive into the geometric significance of these equations and how they apply to practical scenarios.

Let’s start by breaking down the key components of quadratic equations and how to solve them step by step!

EXERCISE 4.2

Q1. Find the roots of the following quadratic equations by factorisation:

(i) x² – 3x – 10 = 0     (ii) 2x² + x – 6 = 0

(iii) √2 x² + 7 x + 5√2 = 0     (iv) 2x² – x + 1/8 = 0

(v) 100x² – 20x + 1 = 0 

Explanation:

1) The quadratic equation is of the form ax² + bx + c = 0, where a ≠ 0.

2) If α and β are the roots of the equation ax² + bx + c = 0, they satisfy the

conditions: 

aα² + bα + c = 0 and aβ² + bβ + c = 0. This means α and β are the solutions to the equation ax² + bx + c = 0.

3) We can solve this quadratic equation ax² + bx + c = 0 using the factorisation

method, which involves breaking the middle term to find the roots.

Solution:

(i) x² – 3x – 10 = 0

1) Quadratic equation: x² – 3x – 10 = 0.

To factorize, note that the constant term is 10, and since its sign is negative, we need factors of 10 whose difference equals – 3.

The factors of 10 are (– 5) and (2), because – 5 + 2 = – 3.

x² – 5x + 2x – 10 = 0

x(x – 5) + 2(x – 5) = 0

(x + 2)(x – 5) = 0

2) So, (x + 2) = 0 or (x – 5) = 0.

3) So, x = – 2 or x = 5.

(ii) 2x² + x – 6 = 0

1) Quadratic equation: 2x² + x – 6 = 0.

To factorize, note that the constant term is 6, and since its sign is negative, we need factors of 6 whose difference equals 1.

The factors of 6 are (– 3) and (4), because – 3 + 4 = 1.

2x² – 3x + 4x – 6 = 0

x(2x – 3) + 2(2x – 3) = 0

(x + 2)(2x – 3) = 0

2) So, (x + 2) = 0 or (2x – 3) = 0

3) So, x = – 2 or x = 3/2.

(iii) √2x² + 7x + 5√2 = 0

1) Quadratic equation: √2x² + 7x + 5√2 = 0

Here last term is 5√2 and its sign is positive. To factorize, we need to break down √2 x 5√2 in such a way

that their sum equals 7.

√2 x 5√2 = 5 x 2, and we need to find factors of 10 whose sum is 7. The factors of 10 are 5 and 2, because 5 + 2 = 7. Now we can use these factors to factorize the equation accordingly.

√2x² + 5x + 2x + 5√2 = 0

x(√2x + 5) + √2(√2x + 5) = 0

(√2x + 5)(x + √2) = 0

2) So, (√2x + 5) = 0, or (x + √2) = 0

3) So, x = (– 5√2/2) or x = – √2.

(iv) 2x² – x + 1/8 = 0

1) Consider the quadratic equation: 2x² – x + 1/8 = 0

2) Multiply the entire equation by 8 to eliminate the fraction: 16x² – 8x + 1 = 0.

The last term is 1, and its sign is positive. We need to factor 16 x 1 in such a way that their sum equals – 8. The factors of 16 that satisfy this condition are 

16 x 1 = (– 4) x (–4) (– 4 – 4 = – 8).

16x² – 4x – 4x + 1 = 0

4x(4x – 1) – (4x – 1) = 0

(4x – 1)(4x – 1) = 0

3) So, (4x – 1) = 0, or (4x – 1) = 0

4) So, x = 1/4 or x = 1/4.

(v) 100x² – 20x + 1 = 0

1)  Quadratic equation: 100x² – 20x + 1 = 0

The last term is 1 and its sign is positive. We need to factor 100 x 1 in such a way that their sum equals – 20. The factors of 100 x 1 = (– 10) x (– 10)

(– 10 – 10 = – 20).

100x² – 10x – 10x + 1 = 0

10x(10x – 1) – (10x – 1) = 0

(10x – 1)(10x – 1) = 0

2) So, (10x – 1) = 0, or (10x – 1) = 0

3) So, x = 1/10 or x = 1/10.

Q2. Solve the problems given in Example 1.

(i) John and Jivanti together have 45 marbles. Both of them lost 5 marbles each, and the product of the number of marbles they now have is 124. We would like to find out how many marbles they had to start with.

(ii) A cottage industry produces a certain number of toys in a day. The cost of production of each toy (in rupees) was found to be 55 minus the number of toys produced in a day. On a particular day, the total cost of production was Rs 750. We would like to find out the number of toys produced on that day.

Explanation:

1) A quadratic equation is of the form ax² + bx + c = 0, where a ≠ 0.

2) Use the given conditions to frame the quadratic equation in this standard form.

3) To solve the quadratic equation ax² + bx + c = 0, apply the factorisation method:

Identify two numbers whose product is $a\times c$ and whose sum is $b$.

Split the middle term using these numbers and factor by grouping.

Solve the resulting linear factors to find the values of $x$.

 

Solution:

(i) John and Jivanti together have 45 marbles. Both of them lost 5 marbles each, and the product of the number of marbles they now have is 124. We would like to find out how many marbles they had to start with.

1) Let the number of marbles John has to be x.

2) According to the problem, Jivanti has (45 – x) marbles.

3) After both lose 5 marbles, John has (x – 5) marbles, and Jivanti has (40 – x)

marbles.

4) Based on the given condition, their current number of marbles satisfies the

equation:

(x – 5)(40 – x) = 124

x(40 – x) – 5(40 – x) = 124
40x – x² – 200 + 5x = 124
– x² + 40x + 5x – 200 – 124 = 0
– x² + 45x – 324 = 0
x² – 45x + 324 = 0

5)  Quadratic equation: x² – 45x + 324 = 0.

The last term is 324 and its sign is positive. We need to factor 324 x 1 in such a way that their sum equals – 45. The factors of 324 x 1 = (– 36) x (– 9)

(– 36 – 9 = – 45).

x² – 36x – 9x + 324 = 0

x(x – 36) – 9(x – 36) = 0

(x – 36)(x – 9) = 0

6) So, (x – 36) = 0, or (x – 9) = 0

7) So, x = 9 or x = 36.

8) John has 36 marbles, and Jivanti has 9.

(ii) A cottage industry produces a certain number of toys in a day. The cost of

production of each toy (in rupees) was found to be 55 minus the number of toys produced in a day. On a particular day, the total cost of production was Rs 750. We would like to find out the number of toys produced on that day.

1) Let the number of toys produced in a day be x.

2) The cost of producing each toy is Rs (55 – x).

3) On a certain day, the total production cost amounted to Rs 750.

4) According to the given condition, the total cost equation can be written as:

x(55 – x) = 750

55x – x² = 750
55x – x² – 750 = 0
– x² + 55x – 750 = 0

x² – 55x + 750 = 0

5)  Quadratic equation: x² – 55x + 750 = 0.

The last term is 750 and its sign is positive. We need to factor 750 x 1 in such a way that their sum equals – 55. The factors of 750 x 1 = (– 25) x (– 30)

(– 25 – 30 = – 55).

x² – 25x – 30x + 750 = 0

x(x – 25) – 30(x – 25) = 0

(x – 25)(x – 30) = 0

6) So, (x – 25) = 0, or (x – 30) = 0

7) So, x = 25 or x = 30.

8) Hence, the number of toys produced that day is 25 or 30.

Q3. Find two numbers whose sum is 27 and whose product is 182.

1) Let the first number be x.

2) The second number is (27 – x).

3) Based on the given condition, the product of these two numbers is 182. Thus, the

equation becomes:

x(27 – x) = 182

27x – x² = 182
27x – x² – 182 = 0
– x² + 27x – 182 = 0

x² – 27x + 182 = 0

4)  Quadratic equation: x² – 27x + 182 = 0.

The last term is 182 and its sign is positive. We need to factor 182 x 1 in such a way that their sum equals – 27. The factors of 182 x 1 = (– 14) x (– 13)

(– 14 – 13 = – 27).

x² – 14x – 13x + 182 = 0

x(x – 14) – 13(x – 14) = 0

(x – 14)(x – 13) = 0

5) So, (x – 14) = 0, or (x – 13) = 0

6) So, x = 13 or x = 14.

7) Hence, the numbers are 13 and 14.

Q4. Find two consecutive positive integers, the sum of whose squares is 365.

1) Let the first positive integer be x.

2) The second consecutive integer will be (x + 1).

3) According to the problem, the sum of their squares is 365, so we can write the

equation:

x² + (x + 1)² = 365

x² + x² + 2x + 1 = 365

2x² + 2x – 365 + 1 = 0

2x² + 2x – 364 = 0 

x² + x – 182 = 0

4)  Quadratic equation: x² + x – 182 = 0.

The last term is 182 and its sign is negative. We need to factor 182 x 1 in such a way that the difference between the factors is 1. The factors of 

182 x 1 = (14) x (– 13)

(14 – 13 = 1).

x² + 14x – 13x – 182 = 0

x(x + 14) – 13(x + 14) = 0

(x + 14)(x – 13) = 0

6) So, (x + 14) = 0, or (x – 13) = 0

7) So, x = –14 or x = 13.

8) As our integer is positive, we have x = 13. 

9) Thus, the two consecutive positive integers are 13 and 14.

Q 5. The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, find the other two sides.

1) Here, a right triangle's altitude depends on its base, so let the base be x.

2) So, altitude will be (x – 7).

3) According to the problem,

x² + (x – 7)² = 13²

x² + x² – 14x + 49 = 13²

2x² – 14x + 49 = 169

2x² – 14x + 49 – 169 = 0

2x² – 14x – 120 = 0

x² – 7x – 60 = 0

4)  Quadratic equation: x² – 7x – 60 = 0.

The last term is 60 and its sign is negative. We need to factor 60 x 1 in such a way that the difference between the factors is – 7. The factors of 

60 x 1 = (12) x (5)

(5 – 12 = – 7).

x² + 5x – 12x – 60 = 0

x(x + 5) – 12(x + 5) = 0

(x + 5)(x – 12) = 0

6) So, (x + 5) = 0, or (x – 12) = 0

7) So, x = – 5 or x = 12.

8) As the sides of a triangle can't be negative, ignore x = – 5, so we have x = 12. 

9) So the base of a triangle is 12 and the altitude is 5.

Q6. 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 Rs 90, find the number of articles produced and the cost of each article.

1) Let the number of pottery articles produced in a day be x.

2) So, the cost of production of each article will be Rs (2x + 3).

3) On one particular day, the total cost was Rs 90.

4) According to the problem,

x(2x + 3) = 90

2x² + 3x = 90

2x² + 3x – 90 = 0

2x² + 3x – 90 = 0

5) Quadratic equation: 2x² + 3x – 90 = 0.

The last term is 90 and its sign is negative. We need to factor 90 x 1 in such a way that the difference between the factors is 3. The factors of 

90 x 1 = (15) x (– 12)

(15 – 12 = 3).

2x² + 15x – 12x + 90 = 0

x(2x + 15) – 6(2x + 15) = 0

(2x + 15)(x – 6) = 0

6) So, (2x + 15) = 0, or (x – 6) = 0

7) So, x = – 15/2 or x = 6.

8) As the number of articles produced can't be negative, so, x can't be – 15/2.

9) So the number of articles produced on that day is 6 and the cost of each

article is Rs 15.

Conclusion: Unlocking the Power of Quadratic Equations

As we conclude our exploration of Quadratic Equations, we’ve unlocked a crucial mathematical tool that helps us solve various complex problems, from calculating areas to predicting trajectories. Mastering the factorization method, completing the square, and applying the quadratic formula equips you with versatile techniques to tackle challenges beyond the classroom. Embrace these concepts, as they lay the groundwork for higher-level math and real-world applications. Keep pushing your limits, and remember, every problem has a solution—just like every quadratic equation has its roots!

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Click here to explore the next step ⇨ NCERT New Syllabus Class 10 - Quadratic Equation Exercise 4.3

Anil Satpute