Our ncert solutions for Exercise 4.3 Class 10 maths 4. Quadratic Equations - ncert solutions - Toppers Study is the best material for English Medium students cbse board and other state boards students.

# Exercise 4.3 Class 10 maths 4. Quadratic Equations - ncert solutions - Toppers Study

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## Exercise 4.3 Class 10 maths 4. Quadratic Equations - ncert solutions - Toppers Study

CBSE board students who preparing for **class 10 ncert solutions maths and Mathematics** solved exercise **chapter 4. Quadratic Equations** available and this helps in upcoming exams
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### You can Find Mathematics solution Class 10 Chapter 4. Quadratic Equations

- All Chapter review quick revision notes for chapter 4. Quadratic Equations Class 10
- NCERT Solutions And Textual questions Answers Class 10 Mathematics
- Extra NCERT Book questions Answers Class 10 Mathematics
- Importatnt key points with additional Assignment and questions bank solved.

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### Exercise 4.3 class 10 Mathematics Chapter 4. Quadratic Equations

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- Exercise 4.3 Class 10 Maths 4. Quadratic Equations - Ncert Solutions - Toppers Study
- Class 10 Ncert Solutions
- Solution Chapter 4. Quadratic Equations Class 10
- Solutions Class 10
- Chapter 4. Quadratic Equations Exercise 4.3 Class 10

## 4. Quadratic Equations

### | Exercise 4.3 |

## Exercise 4.3 Class 10 maths 4. Quadratic Equations - ncert solutions - Toppers Study

**Exercise: 4.3 **

**Q1. Find the roots of the following quadratic equations, if they exist, by the method of completing the square:**

**(i) 2x ^{2} - 7x + 3 = 0 **

**Solution:**a = 2, b = -7, c = 3Now checking for nature of roots,

D = *b ^{2} - 4ac *

D = *(-7) ^{2} - 4 x 2 x 3*

D = *49 - 24 *

D = *25 *

Hence D > 0

∴ There is two different and real roots

2x^{2} - 7x + 3 = 0

Dividing by *a term* 2 we get.

*Putting A term and B term into a ^{2}-2ab+b^{2}*

**(ii) 2x ^{2} + x - 4 = 0;**

**Solution: **

*a = 2, b = 1, c = -4*

Now checking for nature of roots,

D = *b ^{2} - 4ac *

D = *(1) ^{2} - 4 x 2 x -4*

D = *1 - (-32) *

D = *1 + 32 *

D = *33 *

Hence D > 0

∴ There are two distinct and real roots

2x^{2} + x - 4 = 0

**(iii) 4x ^{2} + 4√3x + 3 = 0**

**Solution: **

a =4, b = 4√3, c= 3

D = b^{2} - 4ac = (4√3)^{2} - 4 x 4 x 3

= 48 - 48 = 0

Here D = 0

Therefore, There are two real and equal roots.

4x^{2} + 4√3x + 3 = 0

⇒(2x)^{2} + 2 . 2x. √3 + (√3)^{2} = 0

⇒(2x + √3)^{2}= 0

⇒ (2x + √3)^{ }(2x + √3)^{ }= 0

⇒ 2x + √3 = 0, 2x + √3 = 0

⇒ 2x = - √3, 2x = - √3

⇒ x = √3/2, x = √3/2

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

*a = 2, b = 1, c = 4*

Now checking for nature of roots,

D = *b ^{2} - 4ac *

D = *(1) ^{2} - 4 x 2 x 4*

D = *1 - 32 *

D = *-31 *

Hence D < 0

∴ There is no real root.

∴ Solution cannot be made of 2x^{2} + 1x + 4 = 0

##### Other Pages of this Chapter: 4. Quadratic Equations

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