This set of Class 10 Maths Chapter 2 Multiple Choice Questions & Answers (MCQs) focuses on “Zeros and Coefficients of Polynomial – 2”.
1. What will be the value of (α – β)2, if α and β are the zeros of 4x2 – 27x – 40?
a) \(\frac {1369}{16}\)
b) \(\frac {139}{16}\)
c) \(\frac {1369}{6}\)
d) \(\frac {19}{16}\)
View Answer
Explanation: α and β are the zeros of 4x2 – 27x – 40
α + β = \(\frac {-27}{4}\) and αβ = – 10
(α – β)2 = α2 + β2 – 2αβ
(α – β)2 = (α + β)2 – 2αβ – 2αβ
(α – β)2 = \((\frac {-27}{4})\)2 – 4(-10)
(α – β)2 = \(\frac {729}{16}\) + 40 = \(\frac {1369}{16}\)
2. If β and γ are the zeros of the polynomial ax3 + bx2 + cx + d, then the value of α2 + β2 + γ2 is _________
a) \(\frac {b^2-2c}{a^2}\)
b) \(\frac {b^2-2ca}{a}\)
c) \(\frac {b^2-2ca}{a^2}\)
d) \(\frac {b^2+2ca}{a^2}\)
View Answer
Explanation: β and γ are the zeros of the polynomial ax3 + bx2 + cx + d
So, α + β + γ = \(\frac {-b}{a}\)
αβ + βγ + γα = \(\frac {c}{a}\)
Now, α2 + β2 + γ2 = (α + β + γ)2 – 2(αβ + βγ + γα)
α2 + β2 + γ2 = \((\frac {-b}{a})\)2 – 2\((\frac {c}{a}) = \frac {b^2}{a^2}\) – 2 \(\frac {c}{a} = \frac {b^2-2ca}{a^2}\)
3. The value of αβ + βγ + γα, if α, β and γ are the zeros of 2x3 – 4x2 + 9x – 7 is _______
a) \(\frac {9}{2}\)
b) \(\frac {3}{2}\)
c) \(\frac {9}{8}\)
d) \(\frac {9}{5}\)
View Answer
Explanation: β and γ are the zeros of 2x3 – 4x2 + 9x – 7
The sum of product of two zeros or αβ + βγ + γα = \(\frac {coefficient \, of \, x}{coefficient \, of \, x^3} = \frac {9}{2}\)
4. If α, β and γ are the zeros of 5x3 + 10x2 – x + 20, then the value of αβγ is _______
a) -1
b) 5
c) -10
d) -4
View Answer
Explanation: α, β and γ are the zeros of 5x3 + 10x2 – x + 20
The product of zeros or αβγ = \(\frac {-constant \, term}{coefficient \, of \, x^3} = \frac {-20}{5}\) = -4
5. What will be the polynomial if the value of α + β + γ = -√3 , αβ + βγ + γα = 4 and = \(\frac {-5}{3}\)?
a) x3 + 3x2 + 12x + 5
b) 3x3 + √3 x2 + 4x + 5
c) 3x3 + 3√3 x2 + 12x + 5
d) x3 + √3 x2 + 12x + 5
View Answer
Explanation: α + β + γ = – √3, + βγ + γα = 4, αβγ = \(\frac {-5}{3}\)
∴ f(x) = x3 – (α + β + γ) x2 + (αβ + βγ + γα)x – αβγ
Substituting we get,
f(x) = x3 + √3 x2 + 4x + \(\frac {5}{3}\)
f(x) = 3x3 + 3√3 x2 + 12x + 5
6. If α, β and γ are the zeros of 2x3 – 6x2 + 5x + 2, then the value of α + β + γ is _______
a) 0
b) 1
c) 3
d) 2
View Answer
Explanation: α, β and γ are the zeros of 2x3 – 6x2 + 5x + 2
The sum of zeros or α + β + γ = \(\frac {-coefficient \, of \, x^2}{coefficient \, of \, x^3} = \frac {6}{2}\) = 3
7. If α and β are the zeros of x2 – (5 + 7k)x + 35k, such that α2 + β2 = 172 is then the value of k is _______
a) 6√3
b) √3
c) 3
d) 3√3
View Answer
Explanation: α and β are the zeros of x2 – (5 + 7k)x + 35k
So, α + β = (5 + 7k) and αβ = 35k
Also, α2 + β2 = 172
(α + β)2 – 2αβ = 172
Substituting values we get,
(5 + 7k)2 – 2(35k) = 172
25 + 49k2 + 70k – 70k = 172
49k2 – 147 = 0
Solving we get, k = √3
8. If α and β are the zeros of x2 + (k2 – 1)x – 20, such that α2 – β2 – αβ = 29 and α – β = 9 then, the value of k is _______
a) 1
b) 0
c) 2
d) 3
View Answer
Explanation: α and β are the zeros of x2 + (k2 – 1)x – 20
So, α + β = -(k2 – 1) and αβ = -20
Also, α – β = 9
Now, α2 – β2 – αβ = -11
(α + β)(α – β) – αβ = 29
-(k2 – 1)9 + 20 = 29
-(k2 – 1)9 = 9
-(k2 – 1) = 1
(k2 – 1) = -1
k2 = -1 + 1
k = 0
9. If α and β are the zeros of x2 + bx + c, then the polynomial having -α, -β as zeros is _______
a) -x2 – bx + c
b) x2 + bx + c
c) x2 – bx + c
d) x2 – bx – c
View Answer
Explanation: α and β are the zeros of x2 + bx + c
Also, α + β = -b and αβ = c
Now, -α-β = -(-b) = b and -α × -β = c
Hence, the polynomial with -α, -β as its zeros will be x2 – bx + c
10. If the two zeros of the polynomial x3 – 9x2 -x + 9, are 1 and 9, then the third zero is ________
a) 9
b) 1
c) 2
d) -1
View Answer
Explanation: The given polynomial is x3 – 9x2 – x + 9.
The two zeros of the polynomial are 1 and 9.
We know that, the sum of zeros of the polynomial or α + β + γ = \(\frac {-coefficient \, of \, x^2}{coefficient \, of \, x^3} = \frac {9}{1}\)
1 + 9 + γ = 9
γ = 9 – 10 = -1
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