Test: Zeroes & Coefficients - Class 10 MCQ

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

We know that quadratic equation is of the form x²-(sum of zeros)x+product of zeros
Sum of zeros=2-6=-4
Product of zeros=2*(-6)=-12
x²-(sum of roots )x + product of roots
x²-(-4)x + 12
x²+4x-12
So the equation is x²+4x-12

We have as the roots which means x + and x- are the factors of the quadratic equation. Multiplying x+ and x- and applying a²-b² we get the equation x²-5.

Let one zero of the equation be x
Then, other would be 2x
As we know,
x × 2x = 18
x² = 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.

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.

x² + 7x + 10 = (x + 2)(x + 5)

So, the value of x² + 7x + 10 is zero when x + 2 = 0 or x + 5 = 0

Therefore, the zeroes of x² + 7x + 10 are –2 and –5.

Sum of zeroes = -7 = –(Coefficient of x) / (Coefficient of x²)

Product of zeroes = 10 = Constant term / Coefficient of x²

Sum of zeros= α+β =4
Product of zeros=αβ=4
We have Quadratic equation as
x²-(Sum of zeros)x+Product of zeros
=x²-4x+4

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

x² + 7x + 10 = (x + 2)(x + 5)

So, the value of x² + 7x + 10 is zero when x + 2 = 0 or x + 5 = 0

Therefore, the zeroes of x² + 7x + 10 are –2 and –5.

Sum of zeroes = -7 = –(Coefficient of x) / (Coefficient of x²)

Product of zeroes = 10 = Constant term / Coefficient of x²

Let the polynomial be f(x).

f(x) = x²− (sum of zeros)x +  product of zeros

= x²−2x–15

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

We have quadratic equation of the form
x² - (sum of zeros) x + product of zeros, so comparing this with the given equation we have sum of zeros = -5
So, α + β = -5

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

The given polynomial is

Also, α and β are the zeros of p(x).

and  α + β = α × β

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.

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