PSEB Solutions for Class 10 Maths Chapter 2 Polynomials Ex 2.2
PSEB Solutions for Class 10 Maths Chapter 2 Polynomials Ex 2.2
PSEB 10th Class Maths Solutions Chapter 2 Polynomials Ex 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) Given Quadratic polynomial,
x2 – 2x – 8
∵ [S = -2, P = -8]
= x2 – 4x + 2x – 8
= x (x – 4) + 2 (x – 4)
= (x – 4) (x + 2)
The value of x2 – 2x – 8 is zero,
iff (x – 4) = 0 or (x + 2) = 0
iff x = 4 or x = – 2
Therefore, zeroes of x2 – 2x – 8 are – 2 and 4.
Now, Sum of zeroes = (- 2) + (4) = 2
(iii) Given that, sum of zeroes and products of zeroes of given quadratic polynomial are 0 and √5 respectively.
Let the quadratic polynomial be ax2 + bx + c and its zeroes be α and β
α + β = Sum of zeroes = 0
and αβ = Product of zeroes = √5
Now, ax2 + bx + c = k(x – α) (x – β)
where k is any constant
= k [x2 – (a + (α+ β)x + αβ)
= k[x2 – 0x + √5]
= k[x2 + √5]
for different values of k, we get different quadratic polynomial.
(iv) Given that, sum of zeroes and product of zeroes of given quadratic polynomial are 1 and 1 respectively.
Let the quadratic polynomial be ax2 + bx + c and its zeroes be α and β
α + β = Sum of zeroes = 1 and
αβ = Product of zeroes = 1
Now, ax2 + bx + c = k(x – α) (x – β)
where k is any constant.
= k [x2 – (α + β)x + αβ]
= k [x2 – 0x + √5]
= it [x2 – x + √5]
for different values of k, we get different quadratic polynomial.
(vi) Given that, sum of zeroes and product of zeroes of given quadratic polynomial are 4 and 1 respectively. Let the quadratic polynomial be
ax2 + bx + c and its zeroes be α and β
α + β = sum of zeroes = 4 and
αβ = Product of zeroes = 1
Now, ax2 + bx + c = k(x – α) (x – β) where k is any constant
= k [x2 – (α + β) x + αβ]
= k [x2 – 4x + 1]
For different values of k, we get different quadratic polynomials.
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