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NCERT Class 10 Mathematics Chapter 2 Polynomial
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Polynomial
Chapter – 2
| Exercise 2.1 |
The graphs of v = p(x) are given in fig. 2.10 below, for some polynomials p(x). Find the number of zeroes of p(x), in each case.
Ans: (i) No zeroes.
(ii) 1
(iii) 3
(iv) 2
(v) 4
(vi) 3
| Exercise 2.2 |
1. Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients
(i) X² – 2x – 8
Ans: x²-2x-8=x²-4x+2x-8=x(x-4)+2(x-4) =(x+2)(x-4)
Zeroes are -2 and 4.
Sum of the zeroes = (-2)+(4) = 2 =
-(-2)/1 = -(Coefficient of x) / (Coefficient of x²
Product of the zeroes = ( – 2) (4) = (- 8)/1 = (Coefficient term)/Coefficient of x².
(ii) 4s² – 4 s +1
Ans: 4s² – 4 s +1 = ( 2s – 1)²
∴ The two zeroes are ½, ½
Sum of the zeroes = ½ + ½ = -(-4)/4 = (Coefficient of ×) /(Coefficient of ײ)
Product of two zeroes = (½) (⅓) =¼ (Coefficient)/(coefficient of ײ) .l
(iii) 6ײ-3-7x = 6x²-7x-3 = 6x-9x + 2x-3 = (2x-3) (3x + 1) Putting 2x-3=0 and 3x + 1 = 0
Ans: We get x = 3/2 and x= -1)2 Zeroes of the quadratic polynomial 2 p(x) = 6x²-7x-3
Sum of the two zeroes = (3/2) + (⅓) = 3/2 -⅓ =7/6
= -(-7)/6 = (Coefficient of ×)/ (coefficient of ײ)
Product of the two zeroes = (3/2) × -(⅓) = -½ = (-3/6)
=(Constant term) / (coefficient of ײ)
(iv) 4u² + 8u = 4u (u + 2)
Ans: It gives the two zeroes, 0 and 2 of the polynomial p(u) = 4u² + 8u + u
Sum of the two zeroes 0 + (-) = -2 =(8)/4 = =(Constant term) / (coefficient of ײ)
Product of the two zeroes = (0) (-2) = 0 = (Constant term) / (coefficient of ײ)
(v) t²-15 = t² (√15) =( t²√15) (t-√15)
Ans: If gives the two zeroes of the polynomial p(t) = t² +0t -15 are √15 and √15.
-(Coefficient of t)/Coefficient of t²)
Product of the two zeroes
= (-√15) × (√15) = -15 = (-15) = constant Sum of the two zeroes = Product of the two zeroes (-√15) + √15=0= Coefficient of term)/(Coefficient of t²)
(vi) 3x² – × – 4 3x² = – 4× + 3×- 4
Ans: 3x² – × – 4 3x² = – 4× + 3×- 4
= 3x(× + 1) – 4 (× + 1) = (× + ) ( 3x – 4)
Putting × + 1 = 0 and 3× – 4 = 0, we get = – 1 and × = 4/3
i.e ., the two zeroes of the quadratic polynomial p(×) = 3x² – × – 4 are – 1 and 4/3
Sum of the two zeroes = (-1) + (4/3) = ⅓ = (-1) /3 = (Coefficient of ×) (Coefficient of x²)
Product of the two zeroes = (-) × (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
Ans: Let the polynomial be a x² + b × c and it’s Zeroes be a and b
Here,a × b = ¼ and ab = – 1
Thus the polynomial formed = x² – ( sum of zeroes ) x + product of Zeroes.
= x² – ( ¼) × – 1 = x² – ×/4 – 1
The other polynomial are = k ( x²- ×/4-1)
If k = 4, then polynomial is = 4x²-×-4.
(ii) √2,⅓
Ans: Here a+b = √2 ab=⅓
Thus the polynomial formed = x²( sum of Zeroes) × + product of zeroes.
= x² – (√2) × + ⅓ or x² – (√2) × + ⅓
Other polynomial are k (x² – √2×+⅓)
If k = 3, then the polynomial is 3x² – 3√2 × + 1.
(iii) 0,√5
Ans: Here a + b = 0 and a .b = √5
Thus the polynomial formed formed = x²- ( sum of zeroes) × + product of Zeroes = x² (0) x² (0)x + (√5 = x + √5
(vi) 1,1
Ans: Here a+b = a: b=1 this the polynomial formed
= x² ( sum of Zeroes) × + product of Zeroes = x² – (1) × +1 = x² – × + 1
(v) ¼,¼
Ans: Here a+b = -¼ ab = ¼
Thus the polynomial formed = x² -(sum of zeroes) × product of Zeroes.
= x² – (-¼) × + ¼ = x² -1/4× + ¼
The other polynomial are k (x² + ¼ ×+¼)
If k = 4 ,then the polynomial is 4x²+ × + 1
(vi) 4,1
Ans: Here a+b = ,ab = 1
Thus the polynomial formed = x² – (sum of zeroes) × + product of zeroes.
= x² – (4) × + 1 = x² – 4 × + 1