If one zero of the polynomial 3x -8x+2k +1 is
seven times the other, then find the zeroes and the
value of k.
Answers
Step-by-step explanation:
Correction:-
Polynomial is 3x^2 -8x+2k +1
Given:-
One zero of the polynomial 3x^2 -8x+2k +1 is
seven times the other.
To find:-
Find the zeroes and the value of k ?
Solution:-
Given that
Given quadratic polynomial = 3x^2-8x+2k+1
In comparing this with the standard quadratic polynomial ax^2+bx+c
a = 3
b= -8
c= 2k+1
Let the zeroes of the Polynomial be α and β
α = 7 times the other zero
α =7β
Then the zeroes are 7β and β
now
Sum of the zeores =α +β
α +β = -b/a
=> 7β + β = -(-8)/3
=> 8 β = 8/3
=> β = 1/3 ------------(1)
α = 7β = 7(1/3) = 7/3
α = 7/3-----------(2)
The zeroes are 7/3 and 1/3
And
Product of the zeroes = αβ
=>αβ = c/a
7β×β = (2k+1)/3
=7β^2 = (2k+1)/3
=> 7(1/3)^2 = (2k+1)/3
=> 7(1/9) = (2k+1)/3
=> 7/9 = (2k+1)/3
On applying cross multiplication then
=> 9(2k+1) = 7×3
=> 18k+9 = 21
=> 18k = 21-9
=> 18k = 12
=> k = 12/18
=> k = 2/3
Therefore,k = 2/3
Answer:-
The zeroes of the given polynomial are 7/3 and 1/3
The value of k = 2/3
Used formulae:-
- The standard quadratic polynomial is ax^2+bx+c
- Sum of the zeores =α +β = -b/a
- Product of the zeroes = αβ = c/a
Answer is k=2
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