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
CBSE Class 10 Sample Paper for 2019 Boards
Question 1
Question 2 (Or 1st)
Question 2 (Or 2nd) Important
Question 3 (Or 1st) Important
Question 3 (Or 2nd)
Question 4
Question 5 Important
Question 6 Important
Question 7 (Or 1st)
Question 7 (Or 2nd) Important
Question 8 (Or 1st)
Question 8 (Or 2nd)
Question 9
Question 10 Important
Question 11 Important
Question 12 Important
Question 13 Important
Question 14 You are here
Question 15
Question 16 (Or 1st) Important
Question 16 (Or 2nd)
Question 17 (Or 1st) Important
Question 17 (Or 2nd)
Question 18 Important
Question 19 (Or 1st)
Question 19 (Or 2nd) Important
Question 20
Question 21 (Or 1st) Important
Question 21 (Or 2nd)
Question 22
Question 23 (Or 1st) Important
Question 23 (Or 2nd)
Question 24
Question 25
Question 26
Question 27 (Or 1st)
Question 27 (Or 2nd) Important
Question 28 (Or 1st)
Question 28 (Or 2nd) Important
Question 29
Question 30 Important
CBSE Class 10 Sample Paper for 2019 Boards
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
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