Question

If are the zeroes of the polynomial f(x) = x2 – 5x +3( k -1) such that

− = 11, find the value of k?
pls with full procedure and question is from 10th class​

Answer 1

-7 is ur answer dear

hope it helps u

Answer 2

Step-by-step explanation:

Correct Question:-

If α and β are the zeroes of the Polynomial f(x)=x^2-5x+3(k-1) such that α-β = 11 ,then find the value of k?

Given:-

α and β are the zeroes of the Polynomial f(x)=x^2-5x+3(k-1) such that α-β = 11

To find:-

find the value of k?

Solution:-

Given quadratic polynomial f(x)=f(x)=x^2-5x+3(k-1)

On Comparing this with the standard quadratic Polynomial ax^2+bx+c

a = 1

b=-5

c=3(k-1)

If α and β are the zeroes then

Sum of the zeroes = α + β = -b/a

=>α + β = -(-5)/1

α + β = 5 --------(1)

Given that

α - β = 11 --------(2)

Product of the zeroes = c/a

=>αβ = 3(k-1)/1

αβ = 3(k-1) -------(3)

We know that

(a-b)^2 = (a+b)^2 -4ab

(α - β)^2 = (α + β)^2 -4(αβ)

=>11^2 = 5^2 -4(3(k-1))

=>121 = 25 -12(k-1)

=>121-25 = -12(k-1)

=>96 = -12(k-1)

=>96/-12 = k-1

=>-8 = k-1

=>-8+1 = k

=>-7 = k

=>k=-7

Therefore,k = -7

Answer:-

The value of k for the given problem is -7

Check:-

If k = -7 then f(x)=x^2-5x+3(-7-1)

=>x^2-5x+3(-8)

=>x^2-5x-24

=>x^2-8x+3x-24

=>x(x-8)+3(x-8)

=>(x-8)(x+3)

To get zeroes write f(x)=0

=>(x-8)(x+3) = 0

=>x-8=0 or x+3 = 0

=>x = 8 or x= -3

Let α = 8 and β = -3

α - β = 8-(-3)

=>α - β = 8+3

=>α - β = 11

Verified the given relation.

Used formulae:-

• The standard quadratic Polynomial ax^2+bx+c
• Sum of the zeroes = -b/a
• Product of the zeroes = c/a
• (a-b)^2 = (a+b)^2 -4ab