SAQ for Chapter 2 Polynomials Class 10 Math NCERT
Important Questions1
If zeroes of the polynomial x2 + 4x + 2a are a and 2/a, then find the value of a.
Answer
Product of (zeroes) roots,
or, 2a = 2
Thus a = 1
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2
Find all the zeroes of f(x) = x2 – 2x.
Answer
We have f(x) = x2 – 2x= x(x – 2)
Substituting f(x) = 0, and solving we get x = 0, 2
Hence, zeroes are 0 and 2.
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3
Find the zeroes of the quadratic polynomial √3 x2 – 8x + 4√3.
Answer
We have p(x) = √3x2 – 8x + 4√3= √3x2 – 6x – 2x + 4√3
= √3x (x – 2√3) – 2(x – 2√3)
= (√3x – 2)(x – 2√3)
Substituting p(x) = 0, we have
(√3x – 2) (x – 2√3) p(x) = 0
Solving we get x = 2/√3, 2√3
Hence, zeroes are 2/√3 and 2√3.
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4
Find a quadratic polynomial, the sum and product of whose zeroes are 6 and 9 respectively. Hence find the zeroes.
Answer
Sum of zeroes, α + β = 6Product of zeroes αβ = 9
Now,
p(x) = x2 – (α + β)x + αβ
= x2 – 6x + 9
Thus quadratic polynomial is x2 – 6x + 9.
Now,
p(x) = x2 – 6x + 9
= (x – 3)(x – 3)
Substituting p(x) = 0, we get x = 3, 3
Hence zeroes are 3, 3
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5
Find the quadratic polynomial whose sum and product of the zeroes are 21/8 and 5/16 respectively.
Answer
Sum of zeroes, α + β = 21/8Product of zeroes αβ = 5/16
Now p(x) = x2 – (α + β)x + αβ
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6
Form a quadratic polynomial p(x) with 3 and –2/5 as sum and product of its zeroes, respectively.
Answer
Sum of zeroes, α + β = 3Product of zeroes αβ = –2/5
Now
p(x) = x2 – (α + β)x + αβ
The required quadratic polynomial isÂ
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7
If m and n are the zeroes of the polynomial 3x2 + 11x – 4, find the value of 
 .
 . Answer
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8
If p and q are the zeroes of polynomial f(x) = 2x2 – 7x + 3, find the value of p2 + q2 .
Answer
We have f(x) = 2x2 – 7x + 3
Since, (p + q)2 = p2 + q2 + 2pq
so, p2 + q2 = (p + q)2 – 2pq
Hence p2 + q2 = 37/4.
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9
Find the condition that zeroes of polynomial p(x) = ax2 + bx + c are reciprocal of each other.
Answer
We have p(x) = ax2 + bx + cLet α and 1/ α be the zeroes of p(x), then
Product of zeroes,
So, required condition is c = a
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10
Find the value of k if –1 is a zero of the polynomial p(x) = kx2 – 4x + k.
Answer
We have p(x) = kx2 – 4x + kSince, –1 is a zero of the polynomial, then
p(–1) = 0
k(–1)2 – 4(–1) + k = 0
k + 4 + k = 0
2k + 4 = 0
2k = –4
Hence, k = –2
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11
If α and β are the zeroes of a polynomial x2 – 4√3 x + 3, then find the value of α + β – αβ.
Answer
We have p(x) = x2 – 4√3 x + 3If α and β are the zeroes of x2 – 4√3 x + 3, then
Sum of zeroes,
or, α + β = 4√3
Product of zeroes αβ = c/a = 3/1
or, αβ = 3
Now α + β – αβ = 4√3 – 3 .
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12
Find the values of a and b, if they are the zeroes of polynomial x2 + ax + b.
Answer
We have p(x) = x2 + ax + bSince a and b are the zeroes of polynomial, we get,
Product of zeroes, ab = b ⇒ a = 1
Sum of zeroes, a + b = – a ⇒ b = – 2a = – 2
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13
If α and β are the zeroes of the polynomial f(x) = x2 – 6x + k, find the value of k, such that α2 + β2 = 40.
Answer
We have f(x) = x2 – 6x + k
Now α2 + β2 = (α + β)2 – 2αβ = 40
(6)2 – 2k = 40
36 – 2k = 40
−2k = 4
Thus k = −2
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14
If one of the zeroes of the quadratic polynomial f(x) = 14x2 – 42k2x – 9 is negative of the other, find the value of ‘k’.
Answer
We have f(x) = 14x2 – 42k2x – 9Let one zero be α, then other zero will be –α.
Sum of zeroes α + (−α) = 0.
Thus sum of zero will be 0.
Sum of zeroes,
Thus k = 0.
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15
If one zero of the polynomial 2x2 + 3x + λ be ½ , find the value of λ and the other zero.
Answer
Let, the zero of 2x2 + 3x + λ be ½ and β.
or, β = λ
Hence λ = β = −2
Thus other zero is −2.
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16
If α and β are zeroes of the polynomial f(x) = x2 – x – k, such that α – β = 9, find k.
Answer
We have f(x) = x2 – x – kSince α and β are the zeroes of the polynomial, then
Sum of zeroes,
α + β = 1 …(1)
Given α – β = 9 …(2)
Solving (1) and (2) we get α = 5 and β = −4
or αβ = −k
Substituting α = 5 and β = −4 we have
(5)( −4) = −k
Thus k = 20
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17
If the zeroes of the polynomial x2 + px + q are double in value to the zeroes of 2x2 – 5x – 3, find the value of p and q.
Answer
We have f(x) = 2x2 – 5x – 3Let the zeroes of polynomial be α and β, then
Sum of zeroes α + β = 5/2
Product of zeroes = αβ = −3/2
According to the question, zeroes of x2 + px + q are 2α and 2β.
Sum of zeros, 2α + 2β = −p/1
2(α + β) = −p
substituting α + β = 5/2 we have
or p = −5
Product of zeroes, 2α2β = q/1
4αβ = q
Substituting αβ = −3/2 we have
−6 = q
Thus p = −5 and q = −6.
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18
If α and β are zeroes of x2 – (k – 6)x + 2(2k – 1), find the value of k if α + β = ½αβ.
Answer
We have p(x) = x2 – (k – 6)x + 2(2k – 1)Since α, β are the zeroes of polynomial p(x), we get
α + β = −[−(k – 6)] = k – 6
αβ = 2(2k – 1)
Now α + β = ½αβ
or, k – 6 = 2k – 1
k = −5
Hence the value of k is −5.
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