Find the value of k so that the polynomial x square - 2 + 6 x + 2 into 2 k - 1 has sum of its zeroes two equal half of the product
Answers
Let our polynomial be p(x), then
⇒ p(x) = x² - 2 + 6x + 2(2k - 1)
⇒ p(x) = x² + 6x + 2(- 1) + 2(2k - 1)
⇒ p(x) = x² + 6x + 2(2k - 2)
⇒ p(x) = x² + 6x + 4(k - 1)
Taking zeros of polynomial as α and β,
Given : α + β = αβ/2
⇒ α + β = - Cofficient of x/Cofficient of x²
⇒ α + β = - 6/1 = - 6
⇒ αβ/2 = - 6
⇒ αβ = - 12
Product of zeros = Constant term/Cofficient of x²
⇒ - 12 = 4(k - 1)/1
⇒ - 12 = 4k - 4
⇒ - 12 + 4 = 4k
⇒ - 8/4 = k
⇒- 2 = k
Answer : - 2
Solution :
The given polynomial is x^2 - 6x + 2 x 2k -1
> x^2 - 6x + ( 4k - 1)
The sum of the zeroes of this polynomial is equal to half the product of them.
Sum of zeroes in a polynomial of the form ax^2 + bx + c is -b/a .
Product of zeroes in a polynomial of the form ax^2 + bx + c is c/a .
Thus , this implies that ;
-b/a = 1/2 ( c/a )
> -b/a = c/2a
> -b = c/2
> -2b = c
In the given polynomial
b = -6
c = 4(k - 1)
12 = 4(k - 1)
> k-1 = 3
> k = 4
This is the required value of k .
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