﻿ Class 10 NCERT Math Solution
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TOPICS
Exercise - 2.2

Question-1 :-  Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.
(i) x² – 2x – 8
(ii) 4s² – 4s + 1
(iii) 6x² – 3 – 7x
(iv) 4u² + 8u
(v) t² – 15
(vi) 3x² – x – 4

Solution :-
```(i) p(x) = x² – 2x – 8
= x² – 4x + 2x – 8
= x(x - 4) + 2(x - 4)
= (x + 2)(x - 4)
Zeroes of polynomial :
x + 2 = 0
x = -2
x - 4 = 0
x = 4

Relationship between the zeroes and the coefficients :
p(x) = x² – 2x – 8
a = 1, b = -2, c = -8
Sum of zeroes = α + β = -b/a = -(-2)/1 = 2
Product of zeroes = αβ = c/a = -8/1 = -8
```
```(ii) p(s) = 4s² – 4s + 1
= 4s² – 2s - 2s + 1
= 2s(2s - 1) - 1(2s - 1)
= (2s - 1)(2s - 1)
Zeroes of polynomial :
2s - 1 = 0
2s = 1
s = 1/2, 1/2

Relationship between the zeroes and the coefficients :
p(s) = 4s² – 4s + 1
a = 4, b = -4, c = 1
Sum of zeroes = α + β = -b/a = -(-4)/4 = 1
Product of zeroes = αβ = c/a = 1/4
```
```(iii) p(x) = 6x² – 3 – 7x
= 6x² – 7x - 3
= 6x² – 9x + 2x – 3
= 3x(2x - 3) + 1(2x - 3)
= (2x - 3)(3x + 1)
Zeroes of polynomial :
2x - 3 = 0
2x = 3
x = 3/2
3x + 1 = 0
3x = -1
x = -1/3

Relationship between the zeroes and the coefficients :
p(x) = 6x² – 3 – 7x
a = 6, b = -7, c = -3
Sum of zeroes = α + β = -b/a = -(-7)/6 = 7/6
Product of zeroes = αβ = c/a = -3/6 = -1/2
```
```
(iv) p(u) = 4u² + 8u
= 4u(u + 2)
Zeroes of polynomial :
4u = 0
u = 0
u + 2 = 0
u = -2
Relationship between the zeroes and the coefficients :
p(u) = 4u² + 8u
a = 4, b = 8, c = 0
Sum of zeroes = α + β = -b/a = -(8)/4 = -2
Product of zeroes = αβ = c/a = 0/4 = 0
```
```(v) p(t) = t² – 15
= t² - (√15)²
= (t + √15)(t - √15)
Zeroes of polynomial :
t + √15 = 0
t = -√15
t - √15 = 0
t = √15
Relationship between the zeroes and the coefficients :
p(t) = t² – 15
a = 1, b = 0, c = -15
Sum of zeroes = α + β = -b/a = 0/1 = 0
Product of zeroes = αβ = c/a = -15/1 = -15
```
```(vi) p(x) = 3x² – x – 4
= 3x² – 4x + 3x – 4
= x(3x - 4) + 1(3x - 4)
= (3x - 4)(x + 1)
Zeroes of polynomial :
3x - 4 = 0
3x = 4
x = 4/3
x + 1 = 0
x = -1

Relationship between the zeroes and the coefficients :
p(x) = 3x² – x – 4
a = 3, b = -1, c = -4
Sum of zeroes = α + β = -b/a = -(-1)/3 = 1/3
Product of zeroes = αβ = c/a = -4/3
```

Question-2 :-  Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.
(i) 1/4, -1   (ii) √2, 1/3  (iii) 0, √5  (iv) 1, 1  (v) -1/4, 1/4  (vi) 4, 1.

Solution :-
```(i) 1/4, -1
Given that :
Sum of zeroes = α + β = -b/a = 1/4
Product of zeroes = αβ = c/a = -4/4
Therefore, a = 4, b = -1, c = -4.
In general form, p(x) = ax² + bx + c.
So, quadratic polynomial is 4x² - x - 4.
```
```(ii) √2, 1/3
Given that :
Sum of zeroes = α + β = -b/a = 3√2/3
Product of zeroes = αβ = c/a = 1/3
Therefore, a = 3, b = -3√2, c = 1.
In general form, p(x) = ax² + bx + c.
So, quadratic polynomial is 3x² - 3√2x + 1.
```
```(iii) 0, √5
Given that :
Sum of zeroes = α + β = -b/a = 0/1
Product of zeroes = αβ = c/a = √5/1
Therefore, a = 1, b = 0, c = √5.
In general form, p(x) = ax² + bx + c.
So, quadratic polynomial is x² + √5.
```
```(iv) 1, 1
Given that :
Sum of zeroes = α + β = -b/a = 1/1
Product of zeroes = αβ = c/a = 1/1
Therefore, a = 1, b = -1, c = 1.
In general form, p(x) = ax² + bx + c.
So, quadratic polynomial is x² - x + 1.
```
```(v) -1/4, 1/4
Given that :
Sum of zeroes = α + β = -b/a = -1/4
Product of zeroes = αβ = c/a = 1/4
Therefore, a = 4, b = 1, c = 1.
In general form, p(x) = ax² + bx + c.
So, quadratic polynomial is 4x² + x + 1.
```
```(vi) 4, 1
Given that :
Sum of zeroes = α + β = -b/a = 4/1
Product of zeroes = αβ = c/a = 1/1
Therefore, a = 1, b = -4, c = 1.
In general form, p(x) = ax² + bx + c.
So, quadratic polynomial is x² - 4x + 1.
```
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