**
Question 14
**

Find the zeroes of the following polynomial:

5√5 x
^{
2
}
+ 30x + 8√5

Here, we solve this question by two methods

- Method 1 - Using Splitting the middle term
- Method 2 - Using D Formula

Last updated at Oct. 1, 2019 by Teachoo

**
Question 14
**

Find the zeroes of the following polynomial:

5√5 x
^{
2
}
+ 30x + 8√5

Here, we solve this question by two methods

- Method 1 - Using Splitting the middle term
- Method 2 - Using D Formula

Transcript

Question 14 (Method 1 – Using Splitting the middle term) Find the zeroes of the following polynomial: 5√5 x2 + 30x + 8√5 5√5 x2 + 30x + 8√5 Factorising by splitting the middle term = 5√5 x2 + 20x + 10x + 8√5 = 5√5 x2 + 20x + (5 × 2)x + 8√5 = 5√5 x2 + 20x + (√5 × √5 × 2)x + 8√5 = 5x(√5 x + 4) + √5 × 2(√5 x + 4) = 5x(√5 x + 4) + 2√5 (√5 x + 4) = (5x + 2√5) (√5x + 4) Splitting the middle term method We need to find two numbers whose Sum = 30 Product = 5√5 × 8 √5 = 5 × 8 × √5 × √5 = 40 × 5 = 200 Zeroes are 5x + 2√5 = 0 5x = –2√5 x = (−𝟐√𝟓)/𝟓 √5x + 4 = 0 √5x = –4 x = (−4)/√5 x = (−4)/√5 × √5/√5 x = (−𝟒√𝟓)/𝟓 √5x + 4 = 0 √5x = –4 x = (−4)/√5 x = (−4)/√5 × √5/√5 x = (−𝟒√𝟓)/𝟓 1 mark Question 14 (Method 2 – Using D formula) Find the zeroes of the following polynomial: 5√5 x2 + 30x + 8√5 5√5 x2 + 30x + 8√5 Comparing with ax2 + bx + c a = 5√5, b = 30, c = 8√5 Now, x = (−𝑏 ± √(𝑏^2 − 4𝑎𝑐))/2𝑎 = (−30 ± √(〖(30)〗^2 − 4 × 5√5 × 8√5) )/(2 × 5√5) = (−30 ± √(900 − 4 × 5 × 8 × √5 × √5) )/(10√5) = (−30 ± √(900 − 4 × 5 × 8 × 5))/(10√5) = (−30 ± √(900 −20 × 40))/(10√5) = (−30 ± √(900 − 800))/(10√5) = (−30 ± √100)/(10√5) = (−30 ± 10)/(10√5) 1.5 marks Therefore, x = (−30 + 10)/(10√5) x = (−20)/(10√5) x = (−2)/√5 x = (−2)/√5 × √5/√5 x = (−𝟐√𝟓)/𝟓 x = (−30 − 10)/(10√5) x = (−40)/(10√5) x = (−4)/√5 x = (−4)/√5 × √5/√5 x = (−𝟒√𝟓)/𝟓 . ∴ Zeroes are x = (−𝟐√𝟓)/𝟓, (−𝟒√𝟓)/𝟓 1.5 marks

CBSE Class 10 Sample Paper for 2019 Boards

Paper Summary

Question 1

Question 2 (Or 1st)

Question 2 (Or 2nd)

Question 3 (Or 1st)

Question 3 (Or 2nd)

Question 4

Question 5

Question 6

Question 7 (Or 1st)

Question 7 (Or 2nd)

Question 8 (Or 1st)

Question 8 (Or 2nd)

Question 9

Question 10

Question 11

Question 12

Question 13

Question 14 You are here

Question 15

Question 16 (Or 1st)

Question 16 (Or 2nd)

Question 17 (Or 1st)

Question 17 (Or 2nd)

Question 18

Question 19 (Or 1st)

Question 19 (Or 2nd)

Question 20

Question 21 (Or 1st)

Question 21 (Or 2nd)

Question 22

Question 23 (Or 1st)

Question 23 (Or 2nd)

Question 24

Question 25

Question 26

Question 27 (Or 1st)

Question 27 (Or 2nd)

Question 28 (Or 1st)

Question 28 (Or 2nd)

Question 29

Question 30

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.