# NCERT Solutions For Class 10 Maths Chapter 4 Quadratic Equations Ex 4.4

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Textbook | NCERT |

Board | CBSE |

Category | NCERT Solutions |

Class | Class 10 |

Subject | Maths |

Chapter | Chapter 3 |

Exercise | Class 10 Maths Chapter 4 Quadratic Equations Ex 4.4 |

Number of Questions Solved | 5 |

## NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations Ex 4.4

**NCERT TEXTBOOK EXERCISES**

**Question 1. Find the nature of the roots of the following quadratic equations. If the real roots exist, find them:**

(i) 2x² -3x + 5 = 0

(ii) 3x^{2} – 4√3x + 4 = 0

(iii) 2x^{2}-6x + 3 = 0

**Solution:**

**(i) 2x ^{2} – 3x + 5 = 0**

This is of the form ax

^{2}+ bx + c = 0,

where a = 2, b = -3 and c = 5

Discriminant, D = b

^{2}-4ac

= (-3)

^{2}-4 x 2 x 5 = 9 – 40 = -31

Since, D < 0

Hence, no real roots exist.

**(ii) 3x ^{2} – 4√3x + 4 = 0**

**(iii) 2x ^{2}-6x + 3 = 0**

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

**(1) 2x ^{2} + kx + 3 = 0**

**(2) kx (x – 2) + 6 = 0**

**Solution:**

**Question 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.**

**Solution:**

Let breadth of the rectangular mango grove be x m

Then, the length of rectangular mango grove be 2xm

According to question,

x x 2x = 800

⇒ 2x^{2} = 800

⇒ x^{2} = 400

⇒ x = ±20 [-20 is rejected]

Hence, breadth = 20 m and length = 2 x 20 = 40 m

So, it is possible to design a rectangular mango grove whose length is twice its breadth.

**Question 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.**

**Solution:**

Let the present age of one friend be x years

Then, the present age of other friend be (20 -x) years

4 years ago, one friend’s age was (x – 4) years

4 years ago, other friend’s age was (20 -x – 4) = (16 -x) years

According to question,

(x-4) (16-x) = 48

⇒ 16x – x^{2} – 64 + 4x = 48

⇒ x^{2} – 20x + 112 = 0

This is of the form ax^{2} + bx + c = 0, where, a = 1, b = -20 and c = 112

Discriminant, D = b^{2} – 4ac

= (-20)^{2} – 4 x 1 x 112 = 400 – 448 = – 48 < 0

Since, no real roots exist.

So, the given situation is not possible.

**Question 5. Is it possible to design a rectangular park of perimeter 80 m and area 400 m ^{2}? If so, find its length and breadth.**

**Solution:**

Let the length of rectangular park be x m and breadth be y m.**Given: **area = 400 m^{2}

## NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations Exercise 4.4 PDF

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