Question 14 - 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

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

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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

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