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**NCERT Class 10 Mathematics Chapter 4 Quadratic Equation**

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**Solutions****NCERT Class 10 Mathematics Chapter 4 Quadratic Equation****Quadratic Equation**

**Quadratic Equation****Chapter – 4**

Exercise 4.1 |

**1. Check whether the following are quadratic equations:**

**(i) (x + 1)**^{2}** = 2( x-3)**

Ans: (x + 1)^{2} = 2( x-3)

⇒ x^{2 }+ 2x +1= 2x – 6 ⇒ x^{2 }+7 = 0

⇒ x^{2 }+ 0. X + 7 = 0

It is of the from ax^{2} + bx + c = 0

The given equation is a quadratic equation.

**(ii) (x-2) (x-3) =(x-1) ( x +3)**

Ans: x²-2x=(-2)(3-x)

⇒ x²-2x=-6+2x

⇒ x²- 4x+6= 0

It is of the form ax² + bx + c = 0

∴ The given equation is quadratic.

**(iii) x**^{ }**-2) (x-1) = (x-1) = (x-1) (x + 3)**

Ans: (x-2) (x+1) = (x-1) (x+3)

⇒ x²+x-2x-2=x²+3x-x-3

⇒ x-x-2=x²+2x-2x-3

⇒ -x-2x-2+3=0

⇒ -3x+1=0 or 3x-1=0

It is not of the form ax² + bx + c = 0

∴ The given equation is not quadratic equation.

**(iv) (x – 3) (2x +1) = x (x+5)**

Ans: (x-3) (2x+1)=x(x+5)

2x² + x-6x – 3 = x² + 5x

→ 2-5x-3 = x² + 5x

2 -x- 5x – 5x – 2 = 0

⇒ -10x-3=0 It is a quadratic equation.

**(v) (2x – 1) (x-3) = (x+5) (x-1)**

Ans: (x-3) (2x+1)=x(x+5)

2x² + x-6x – 3 = x² + 5x

→ 2 – 5x -3 = x² + 5x

2 -x -5x – 5x – 2 = 0

⇒ -10x-3=0 It is a quadratic equation.

**(vi) x**^{2}** +3x + 1 = (x-)**^{2}

Ans: x²+3 x +1=(x-2)

X² + 3x + 1= x² – 4x + 4

3x + 4x + 1-4=0

⇒ 7x-3=0 It is a linear equation and not a quadratic equation.

**(vii) (x-3)**^{3}** = 2x (x**^{2 }**-1 )**

Ans: (x+2)² = 2x (x² – 1)

⇒ x² + 3x² x 2 + 3x x 2² + 23 = 2x² – 2x

⇒ x + 6x² + 12x + 8 = 2x²-2x

⇒ -x + 6x² + 14x + 8 = 0

⇒ x²-6x² – 14x – 8 = 0

It is a cubic equation and not a: quadratic equation.

**(viii) x3 – 4 x2 – x +1 = (x- 2)**^{3}

Ans: x²-4x²-x+1=(x-2)

⇒ -4x² – x + 1 = x² + 3(x)(-2)(x-2)-2

⇒ x² – 4x² – x + 1 = x² – 6x(x-2)-8

⇒ x-4x²-x + 1 = x²-6x² + 12x-8

→ 4x² + 6x² – x – 2x + 1 + 8 = 02-13x + 9 = 0 It is a quadratic equation.

**2. Represent the following situations in the form of quadratic equation:**

**(i) The area of a rectangular plot is 528 m². 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.**

Ans: Let breadth be = x metres

Then length =(2x+1) metres

X x (2x+1) = 528 (Area of the plot)

**Or**

2x² + x – 528 = 0

**(ii) The product of two consecutive positive integers is 306. We need to find the integers.**

Ans: Let the positive consecutive integers be x and x + 1 We are given that x x (x + 1) = 306 x² + x-306 = 0

**(iii) Rohan’s mother is 26 years older than him. The product of their ages (in years) 3 years from will be 360. We would like to find Rohan’s present age.**

Ans: Rohan’s present age = x years Then present age of Rohan’s mother = (x + 26) years.

After three years, its is given that,

( x+3) x [(x+26) + 3] = 360

(x+3) x (x+29) = 360

X² + 32 x+87 = 360

**Or**

x² + 32x + 87-360 = 0 x² + 32x – 273 = 0

**(iv) A train travels a distance of 480 km. at a uniform speed. If the speed had been 8km/h less, then it would have taken 3 hours more to cover the same distance. We need to find the speed of the train. Solution: (i) Let breadth be = x metres**

Ans: Do yourself.

Exercise 4.2 |

**1. Find the roots of the following quadratic equations Factorisation:**

**(i) x²-3x-10 = 0**

Ans: x²-3x-100x²-5x+2x-10=0

⇒ x(x-5)+2(x – 5) = 0

⇒ (x-5) (x + 2) = 0

⇒ (x-5) = 0 or (x + 2) = 0

⇒ x = 5 or x =-2

Hence, the two roots are -2 and 5

**(ii) 2x ^ 2 + x – 6 = 0**

Ans: 2x²+x-6=02x²+4x-3x-6=0

2x (x+2)-3 (x+2) = 0

(x+2) (2x-3)=0

⇒ X+2 = 0 or 2x-3 = 0

⇒ x = -2 or x = 2

**(iii) √2x² + 7x +5√2=0**

Ans: √2x² + 7x 5√2=0

√2x² +5+2x +5√2 = 0

x(√2x +5 + √2(√2 +5) = 0

(x + √2x) (√2 +5) =0

**(iv) 2x**^{2 }**– x + 1/8 = 0**

Ans: Do yourself.

**(v) 100x²-20x + 1 = 0**

Ans: Do yourself.

**2. Represent the following situations mathematically:**

**(i) John Jivanti together has 45 marbles. Both of them lost 5 marbles each, and the product of the number of marbles they now have is 324. We would like to find out how many marbles they had to start with.**

Ans: The quadratic equation is * x ^ 2 – 45x + 324 = 0 Where x is the number of marbles with John. The equation can be rewritten as

X² – 36x – 9x + 324 = 0

⇒ x(x-36)-9(x-36) = 0

⇒ (x-9) (x−36) = 0 ⇒ x = 9 or 36

Therefore, either john had 9 marbles and jivanti had 36 marbles or vice versa

**(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.**

Ans: The quadratic equation is x² – 55x + 750 = 0 … (i) Where x is the number of toys produced on that day. (i) can be rewritten as x²-30x25x + 750 = 0

⇒ x(x-30)-25(x – 30) = 0

⇒ (x-25) (x-30) = 0 ⇒ x=25 or x = 30

Hence, the cost of production for the production of 25 toys is the same as in the case when 30 toys are made.

**3. Find two numbers whose sum is 27 and product is 182.**

Ans: Let one number be x, then the second number = 27 – x According to question,

Hence, the two numbers are 13 and 14

**4. Find two consecutive positive integers, sum of whose squares is 365.**

Ans: Let the consecutive positive integers be x and x + 1 We are given that

**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.**

Ans: In ΔΑΒC, base BC and altitude AC = x cm. = (x-7) cm. ∠ACB = 90°, AB = 13cm.

By Pythagoras Theorem, we have, BC2 + AC2 = AB2 ⇒ x² + (x – 7)² = 132

⇒ x²+x² – 14x + 49 = 169

2x²-14x – 120 = 0

⇒ x²-7x – 60 = 0

⇒ x² – 12x + 5x-60 = 0

⇒ x(x-12) + 5(x – 12) = 0

⇒ 1 – 5x – 12

Sides are not negative 12 2.BC = 12 cm and AC = 5cm.

**6. A cottage industry produces a certain number of pottery articles in a day. It was observed on a particular day 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.**

Ans: Let the number of articles produced in a day (on a particular day )= x

x(2x + 3) = 90

We are given that,

⇒ 2x² + 3x – 90 = 0

⇒ 2x² + 15x-12x – 90 = 0

⇒ x(2x+15) – 6(2x + 15) = 0

⇒ (2x+15) (x-6) = 0

Now v, 2x + 15 =0 Rightarrow 2x=-15 Or, x -6=0 Rightarrow x = 6 15 x=-2 (not exist)

Hence, the number of articles = 6

Cost of production of one article = Rs.(2 x 6 + 3) = Rs. 15

Exercise 4.3 |

**1. Find the roots of the following quadratic equations, if they exist, by the method of completing the square:**

**(i) 2x² – 7x + 3 = 0**

Ans:

**(ii) 2x**^{2}** 2 + x – 4 = 0**

Ans:

** (iii) 4x**^{2}** +4√3x + 3 = 0**

Ans:

**(iv) 2x**^{2}**2 + x + 4 = 0 **

Ans:

**(v) 100x**^{2}** – 3x – 10 = 0**

Ans:

**2. Find the values of k for each of the following quadratic equations, so that they have two equal roots.**

**(i) 2x**^{2}** – 3 x + 5 = 0**

Ans: Do yourself.

**(ii) kx (x – 2) + 6 = 0**

Ans: Do yourself.

**3. Is it possible to design a rectangular mango grove whose length is twice its breadth,and the area is 800 m**^{2}**? If so, find its length and breadth.**

Ans: let breadth be x, then will be 2x

= Area of rectangular mango grove = 800 m^{2}

= x × 2x = 800

2x^{2} = 800

X^{2} = 400

∴ x = 20

= x can not be negative

∴ x = 20

= Breadth of the rectangular mango grove = x = 20 m

= Length of the rectangular mango grove = 2 × 20 × 2 = 40

**4. 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 ages in years was 48.**

Ans: Let their present ages be and 20 – x yrs respectively

Four years ago, their ages would be (x – 4) and (16 – x) respectively.

( x – 4) (16 + x) = 48

16 x – x^{2} – 6 4 + 4 x = 48

Since the discrimination, < o, the roots would be imaginary.

Hence, such a situation is not possible.

**5. Is it possible to design a rectangular park of perimeter 80 m and area 400 m2? If so, find its length and breadth.**

Ans: Do yourself.