Mathematics Questions and Answers – Zeros and Coefficients of Polynomial – 2

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This set of Mathematics Quiz for Class 10 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

Answer: a
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 = \(\frac {1369}{16}\)
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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

Answer: c
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

Answer: a
Explanation: β and γ are the zeros of 2x3 – 4x2 + 9x – 7
The sum of product of two zeros or αβ + βγ + γα = \(\frac {coefficient \, of \, x}{coefficient \, of \, x^3} = \frac {9}{2}\)
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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

Answer: d
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

Answer: c
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
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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

Answer: c
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

Answer: b
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
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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

Answer: b
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

Answer: c
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
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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

Answer: d
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

Sanfoundry Global Education & Learning Series – Mathematics – Class 10.

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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn | Youtube | Instagram | Facebook | Twitter