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QUESTION: 1

If α and β are the zeroes of the polynomial 5x^{2} – 7x + 2, then sum of their reciprocals is:

Solution:

We have 2 find (1/α + 1/β)

now 1/α + 1/β = (α + β)/ α β (taking LCM)

now by the given poly. we get

(α + β) = -b/a = 7/5

α β = c/a = 2/5

so, (α + β)/ α β = (7/5) / (2/5)

= 7/2

So, 1/α + 1/β = (α + β)/ α β = 7/2

Hence, 1/α + 1/β = 7/2

QUESTION: 2

If one zero of polynomial f(x) = (k^{2} + 4)x^{2} + 13x + 4k is reciprocal of the other, then k =

Solution:

QUESTION: 3

Find the quadratic polynomial whose zeros are 2 and -6

Solution:

We know that quadratic equation is of the form x^{2}-(sum of zeros)x+product of zeros

Sum of zeros=2-6=-4

Product of zeros=2*(-6)=-12

So the equation is x^{2}+4x-12

QUESTION: 4

If α, β, γ be the zeros of the polynomial p(x) such that α + β + γ = 3, αβ + βγ + γα = 10 and αβγ = -24, then p(x) is

Solution:

QUESTION: 5

If -√5 and √5 are the roots of the quadratic polynomial. Find the quadratic polynomial.

Solution:

We have as the roots which means x + and x- are the factors of the quadratic equation. Multiplying x+ and x- and applying a^{2}-b^{2 }we get the equation x^{2}-5.

QUESTION: 6

If one root of polynomial equation ax^{2} + bx + c = 0 be reciprocal of other, then

Solution:

QUESTION: 7

If one zero of the polynomial x^{2} + kx + 18 is double the other zero then k = ?

Solution:

Let one zero of the equation be x

Then, other would be 2x

As we know,

x × 2x = 18

x^{2} = 9

x = 3, -3

Case 1:

If x = 3, 2x = 6

Therefore, -k = sum of roots = 9

⇒ k = -9

Case 2:

If x = -3, 2x = -6

Therefore, -k = -9

⇒ k = 9

So, k can be 9,-9. Therefore, option 'b' is correct.

QUESTION: 8

If one zero of 2x^{2} – 3x + k is reciprocal to the other, then the value of k is :

Solution:

Given: 2x²-3x+k .............eq 1

Let the 2 zeroes be α & 1/α

Quadratic form : ax²+bx+c.........eq 2

On comparing eq 1 & eq 2 we get

a=2 ,b = -3, c= k

Product of zeroes = α × 1/α = 1

Product of zeroes= c/a

1= k/ 2

K= 2

The value of k is 2.

QUESTION: 9

Sum and the product of zeroes of the polynomial x^{2} +7x +10 is

Solution:

x^{2} + 7x + 10 = (x + 2)(x + 5)

So, the value of x^{2} + 7x + 10 is zero when x + 2 = 0 or x + 5 = 0

Therefore, the zeroes of x^{2} + 7x + 10 are –2 and –5.

Sum of zeroes = -7 = –(Coefficient of x) / (Coefficient of x^{2})

Product of zeroes = 10 = Constant term / Coefficient of x^{2}

QUESTION: 10

If sum of the zeroes of the polynomial is 4 and their product is 4, then the quadratic polynomial is

Solution:

Sum of zeros= α+β =4

Product of zeros=αβ=4

We have Quadratic equation as

x^{2}-(Sum of zeros)x+Product of zeros

=x^{2}-4x+4

QUESTION: 11

What value/s can x take in the expression k(x – 10) (x + 10) = 0 where k is any real number.

Solution:

k(x – 10) (x + 10) =0

⇒ either k=0

Or x-10=0

Or x+10=0

Since we don’t know the value of k

So either x-10=0

x=10

Or x+10=0

x=-10

So values of x can be 10,-10

QUESTION: 12

Find the sum and the product of zeroes of the polynomial x^{2} +7x +10

Solution:

x^{2} + 7x + 10 = (x + 2)(x + 5)

So, the value of x^{2} + 7x + 10 is zero when x + 2 = 0 or x + 5 = 0

Therefore, the zeroes of x^{2} + 7x + 10 are –2 and –5.

Sum of zeroes = -7 = –(Coefficient of x) / (Coefficient of x^{2})

Product of zeroes = 10 = Constant term / Coefficient of x^{2}

QUESTION: 13

The sum and product of zeros of a quadratic polynomial are 2 and -15 respectively. The quadratic polynomial is

Solution:

QUESTION: 14

If the product of two zeros of the polynomial f(x) = 2x^{3} + 6x^{2} – 4x + 9 is 3, then its third zero is

Solution:

Step-by-step explanation:

p(x) = 2x³ + 6x² – 4x + 9

Define α,β and γ

Let α,β and γ be the 3 roots

αβ = 3 (Given)

Find the products of all the zeros:

αβγ = - d/a

αβγ = - 9/2

Find the third zeros:

Given αβ = 3

3γ = -9/2

γ = -9/2 ÷ 3

γ = -3/2

The third zero is -3/2

QUESTION: 15

If α, β are zeroes of the polynomial f(x) = x^{2} + 5x + 8, then value of (α + β) is

Solution:

We have quadratic equation of the form

x^{2 }- (sum of zeros) x + product of zeros, so comparing this with the given equation we have sum of zeros = -5

So, α + β = -5

QUESTION: 16

If sum of the squares of zeros of the quadratic polynomial f(x) = x^{2} – 8x + k is 40, find the value of k.

Solution:

p (x)= x^2-8x+k

p (x)=Ax^2+Bx+C(The equation is in this form)

Let the zeroes be 'a' and 'b'

It is given that

=> a^2+b^2=40

=> (a+b)^2 - 2ab = 40 ----1

sum of zeroes

=> a+b = -B/A = -(-8)/1 = 8

Product of zeroes

=> a*b = C/A = k

Substitute these values in the equation1 we get

=> 8^2 - 2k = 40

=> 64 - 2k = 40

=> 2k = 24

=> k = 12

Therefore the value of k is 12

QUESTION: 17

If α , β are the zeroes of f(x) = px^{2} – 2x + 3p and α + β = αβ then the value of p is:

Solution:

The given polynomial is

Also, α and β are the zeros of *p*(*x*).

and α + β = α × β

QUESTION: 18

If the two zeroes of the quadratic polynomial 7x^{2} – 15x – k are reciprocals of each other, the value of k is:

Solution:

QUESTION: 19

The number of polynomials having zeroes -2 and 5 is:

Solution:

Since the question doesn’t say that there are only 2 zeros of the polynomial we , there can be n number of polynomials which have two of its zeros as -2 and 5 .So the correct answer is more than 3.

QUESTION: 20

Find the sum and the product of the zeroes of the polynomial: x^{2}-3x-10

Solution:

X²-3x-10

x² -(5x-2x)-10

x² - 5x+2x-10

x(x-5)+2(x-5)

(x-5)(x+2)

x=5

x=-2

Sum of zeroes = α+β = 5-2 = 3

α+β = -b/a = -(-3)/1 = 3

Product of zeroes = αβ = 5*-2 = -10

αβ = c/a = -10/1 = -10

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