Find the value of k such that the polynomial:x2–(k + 6)x+2(2k -1) has the sum of its zeros equal to half of their products.
Answers
Step-by-step explanation:
Given:-
A polynomial:x²–(k + 6)x+2(2k -1) has the sum of its zeros equal to half of their products.
To find:-
Find the value of k ?
Solution:-
Given quadratic polynomial is x²–(k + 6)x+2(2k -1)
On comparing with the standard quadratic polynomial ax²+bx+c then
a = 1
b = -(k+6)
c = 2(2k-1)
We know that
Sum of the zeroes = -b/a
=> -[-(k+6)]/1
=> k+6
The sum of the zeroes = k+6 -------------(1)
Product of the zeroes = c/a
=> 2(2k-1)/1
=> 2(2k-1)
The product of the zeroes = 2(2k-1) ------(2)
According to the given problem
The sum of the zeores is equal to the half of the product of the zeroes
From (1)&(2)
=> k+6 = 2(2k-1)/2
=>k+6 = 2k-1
=> 2k-1 = k+6
=> 2k-k = 6+1
=> k = 7
Therefore, k = 7
Answer:-
The value of k for the given problem is 7
Check:-
If k = 7 then the Polynomial becomes
x²-(7+6)x+2(2(7)-1))
=> x²-13x+26
Sum of the zeroes = 13
Product of the zeroes = 26
Sum of the zeroes = 13
= 26/2
= Product of the zeroes/2
The sum of the zeores is equal to the half of the product of the zeroes
Verified the given relations in the given problem
Used formulae:-
- Thee standard quadratic polynomial is ax²+bx+c
- Sum of the zeroes = -b/a
- Product of the zeroes = c/a