NCERT Solutions 10th Maths Chapter 1 Polynomials Exercise 1.2

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Exercise 2.2
1. Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.
(i) x2 ñ 2x ñ 8
(ii) 4s2 ñ 4s + 1
(iii) 6x2 ñ 3 ñ 7x
(iv) 4u2 + 8u
(v) t2 ñ 15
(vi) 3x2 ñ x ñ 4
Answer
(i) x2 ñ 2x ñ 8
= (x - 4) (x + 2)
The value of x2 ñ 2x ñ 8 is zero when x - 4 = 0 or x + 2 = 0, i.e., when x = 4 or x = -2
Therefore, the zeroes of x2 ñ 2x ñ 8 are 4 and -2.
Sum of zeroes = 4 + (-2) = 2 = -(-2)/1 = -(Coefficient of x)/Coefficient of x2
Product of zeroes = 4 ◊ (-2) = -8 = -8/1 = Constant term/Coefficient of x2
(ii) 4s2 ñ 4s + 1
= (2s-1)2
The value of 4s2 - 4s + 1 is zero when 2s - 1 = 0, i.e., s = 1/2
Therefore, the zeroes of 4s2 - 4s + 1 are 1/2 and 1/2.
Sum of zeroes = 1/2 + 1/2 = 1 = -(-4)/4 = -(Coefficient of s)/Coefficient of s2
Product of zeroes = 1/2 ◊ 1/2 = 1/4 = Constant term/Coefficient of s2.
(iii) 6x2 ñ 3 ñ 7x
= 6x2 ñ 7x ñ 3
= (3x + 1) (2x - 3)
The value of 6x2 - 3 - 7x is zero when 3x + 1 = 0 or 2x - 3 = 0, i.e., x = -1/3 or x = 3/2
Therefore, the zeroes of 6x2 - 3 - 7x are -1/3 and 3/2.
Sum of zeroes = -1/3 + 3/2 = 7/6 = -(-7)/6 = -(Coefficient of x)/Coefficient of x2
Product of zeroes = -1/3 ◊ 3/2 = -1/2 = -3/6 = Constant term/Coefficient of x2.
(iv) 4u2 + 8u
= 4u2 + 8u + 0
= 4u(u + 2)
The value of 4u2 + 8u is zero when 4u = 0 or u + 2 = 0, i.e., u = 0 or u = - 2
Therefore, the zeroes of 4u2 + 8u are 0 and - 2.
Sum of zeroes = 0 + (-2) = -2 = -(8)/4 = -(Coefficient of u)/Coefficient of u2
Product of zeroes = 0 ◊ (-2) = 0 = 0/4 = Constant term/Coefficient of u2.
(v) t2 ñ 15
= t2 - 0.t - 15
= (t - v15) (t + v15)
The value of t2 - 15 is zero when t - v15 = 0 or t + v15 = 0, i.e., when t = v15 or t = -v15
Therefore, the zeroes of t2 - 15 are v15 and -v15.Sum of zeroes = v15 + -v15 = 0 = -0/1 = -(Coefficient of t)/Coefficient of t2
Product of zeroes = (v15) (-v15) = -15 = -15/1 = Constant term/Coefficient of u2.
(vi) 3x2 ñ x ñ 4
= (3x - 4) (x + 1)
The value of 3x2 ñ x ñ 4 is zero when 3x - 4 = 0 and x + 1 = 0,i.e., when x = 4/3 or x = -1
Therefore, the zeroes of 3x2 ñ x ñ 4 are 4/3 and -1.
Sum of zeroes = 4/3 + (-1) = 1/3 = -(-1)/3 = -(Coefficient of x)/Coefficient of x2
Product of zeroes = 4/3 ◊ (-1) = -4/3 = Constant term/Coefficient of x2.
2. Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.
(i) 1/4 , -1
(ii) v2 , 1/3
(iii) 0, v5
(iv) 1,1
(v) -1/4 ,1/4
(vi) 4,1
Answer
(i) 1/4 , -1
Let the polynomial be ax2 + bx + c, and its zeroes be a and fl
a + fl = 1/4 = -b/a
afl = -1 = -4/4 = c/a
If a = 4, then b = -1, c = -4
Therefore, the quadratic polynomial is 4x2 - x -4.
(ii) v2 , 1/3
Let the polynomial be ax2 + bx + c, and its zeroes be a and fl
a + fl = v2 = 3v2/3 = -b/a
afl = 1/3 = c/a
If a = 3, then b = -3v2, c = 1
Therefore, the quadratic polynomial is 3x2 -3v2x +1.
(iii) 0, v5
Let the polynomial be ax2 + bx + c, and its zeroes be a and fl
a + fl = 0 = 0/1 = -b/a
afl = v5 = v5/1 = c/a
If a = 1, then b = 0, c = v5
Therefore, the quadratic polynomial is x2 + v5.
(iv) 1, 1
Let the polynomial be ax2 + bx + c, and its zeroes be a and fl
a + fl = 1 = 1/1 = -b/a
afl = 1 = 1/1 = c/a
If a = 1, then b = -1, c = 1
Therefore, the quadratic polynomial is x2 - x +1.
(v) -1/4 ,1/4
Let the polynomial be ax2 + bx + c, and its zeroes be a and fl
a + fl = -1/4 = -b/a
afl = 1/4 = c/a
If a = 4, then b = 1, c = 1
Therefore, the quadratic polynomial is 4x2 + x +1.
(vi) 4,1
Let the polynomial be ax2 + bx + c, and its zeroes be a and fl
a + fl = 4 = 4/1 = -b/a
afl = 1 = 1/1 = c/a
If a = 1, then b = -4, c = 1
Therefore, the quadratic polynomial is x2 - 4x +1.