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 4x^{2} – 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 4x

^{2}– 27x – 40

α + β = \(\frac {-27}{4}\) and αβ = – 10

(α – β)

^{2}= α

^{2}+ β

^{2}– 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 ax^{3} + bx^{2} + cx + d, then the value of α^{2} + β^{2} + γ^{2} 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 ax

^{3}+ bx

^{2}+ cx + d

So, α + β + γ = \(\frac {-b}{a}\)

αβ + βγ + γα = \(\frac {c}{a}\)

Now, α

^{2}+ β

^{2}+ γ

^{2}= (α + β + γ)

^{2}– 2(αβ + βγ + γα)

α

^{2}+ β

^{2}+ γ

^{2}= \((\frac {-b}{a})\)

^{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 2x^{3} – 4x^{2} + 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 2x

^{3}– 4x

^{2}+ 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 5x^{3} + 10x^{2} – 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 5x

^{3}+ 10x

^{2}– 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) x^{3} + 3x^{2} + 12x + 5

b) 3x^{3} + √3 x^{2} + 4x + 5

c) 3x^{3} + 3√3 x^{2} + 12x + 5

d) x^{3} + √3 x^{2} + 12x + 5

View Answer

Explanation: α + β + γ = – √3, + βγ + γα = 4, αβγ = \(\frac {-5}{3}\)

∴ f(x) = x

^{3}– (α + β + γ) x

^{2}+ (αβ + βγ + γα)x – αβγ

Substituting we get,

f(x) = x

^{3}+ √3 x

^{2}+ 4x + \(\frac {5}{3}\)

f(x) = 3x

^{3}+ 3√3 x

^{2}+ 12x + 5

6. If α, β and γ are the zeros of 2x^{3} – 6x^{2} + 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 2x

^{3}– 6x

^{2}+ 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 x^{2} – (5 + 7k)x + 35k, such that α^{2} + β^{2} = 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 x

^{2}– (5 + 7k)x + 35k

So, α + β = (5 + 7k) and αβ = 35k

Also, α

^{2}+ β

^{2}= 172

(α + β)

^{2}– 2αβ = 172

Substituting values we get,

(5 + 7k)

^{2}– 2(35k) = 172

25 + 49k

^{2}+ 70k – 70k = 172

49k

^{2}– 147 = 0

Solving we get, k = √3

8. If α and β are the zeros of x^{2} + (k^{2} – 1)x – 20, such that α^{2} – β^{2} – 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 x

^{2}+ (k

^{2}– 1)x – 20

So, α + β = -(k

^{2}– 1) and αβ = -20

Also, α – β = 9

Now, α

^{2}– β

^{2}– αβ = -11

(α + β)(α – β) – αβ = 29

-(k

^{2}– 1)9 + 20 = 29

-(k

^{2}– 1)9 = 9

-(k

^{2}– 1) = 1

(k

^{2}– 1) = -1

k

^{2}= -1 + 1

k = 0

9. If α and β are the zeros of x^{2} + bx + c, then the polynomial having -α, -β as zeros is _______

a) -x^{2} – bx + c

b) x^{2} + bx + c

c) x^{2} – bx + c

d) x^{2} – bx – c

View Answer

Explanation: α and β are the zeros of x

^{2}+ 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 x

^{2}– bx + c

10. If the two zeros of the polynomial x^{3} – 9x^{2} -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 x

^{3}– 9x

^{2}– 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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