Answer:
[tex]\sf k = \dfrac{-2}{3}[/tex]
Step-by-step explanation:
kx² + 2x + 3k = 0
a = k; b = 2 and c = 3k
Now, we can find the sum of zeroes.
[tex]\sf \alpha + \beta = \dfrac{-b}{a}\\\\[/tex]
[tex]\sf = \dfrac{-2}{k}[/tex]
Find the product of zeroes.
[tex]\sf \alpha *\beta = \dfrac{c}{a}[/tex]
[tex]\sf = \dfrac{3k}{k}\\\\ = 3[/tex]
It is given that sum of zeroes is equal to product of zeroes.
[tex]\sf \alpha + \beta = \alpha *\beta[/tex]
[tex]\sf \dfrac{-2}{k}=3\\\\Cross \ multiply,\\\\ -2 = 3k\\\\ \dfrac{-2}{3}=k\\[/tex]
[tex]\boxed{\sf k = \dfrac{-2}{3}}[/tex]
The value of k for the quadratic polynomial when the sum of the zeroes equals their product is -2/3. This is obtained by applying Vieta's formulas to the given polynomial and solving for k.
To find the value of k for the quadratic polynomial kx2+2x+3k when the sum of the zeroes is equal to their product, we utilize Vieta's formulas. For a quadratic equation ax2+bx+c=0, the sum of the zeroes (let's call them p and q) is -b/a, and the product of the zeroes is c/a. Applying this to our polynomial where a=k, b=2, and c=3k, we get the following system of equations:
Since the sum is equal to the product:
-2/k = 3
Multiplying both sides by k to solve for k:
-2 = 3k
k = -2/3
So, the value of k is -2/3 for the given conditions.
