Question

find k if sum of the zeroes of the polynomial x²- (k+6) x+2 (2k+1) is equal to half of their product​.


50 Point is here so no spam And no wrong answer

Answer 1

Answer:

find k if sum of the zeroes of the polynomial x²- (k+6) x+2 (2k+1) is equal to half of their product

Answer 2

QuestioN :

find k if sum of the zeroes of the polynomial x²- (k+6) x+2 (2k+1) is equal to half of their product​.

ANswer :

The value of k is 5.

SolutioN :

The given quadratic polynomial is x² - ( k + 6 ) x + 2 ( 2k + 1 ).

We have to find the value of k.

Now,

x² - ( k + 6 ) x + 2 ( 2k + 1 )

Comparing with ax² + bx + c = 0, we get,

• a = 1

• b = - ( k + 6 )

• c = 2 ( 2k + 1 )

Now,

$Sum \:\:of\:\:zeroes\:(\alpha +\beta )\:=\:-\frac{b}{a}$

$=\:\alpha +\beta \:=\:\frac{-(K+6)}{1}$

$=\:\alpha +\beta \:=\:-(-K+6)$

$=\:\alpha +\beta =\:K+6--(1)$

Now,

$Product \:\:of \:\:zeroes(\alpha +\beta )=\frac{c}{a}$

$=\alpha .\beta =\frac{2(2k+1)}{1}$

$=\alpha .\beta =4k+2--(2)$

Now,

$\alpha +\beta =\frac{\alpha .\beta }{2}---\:\:[Given]$

$=k+6=\frac{4k+2}{2} ---[From\:(1)\:and\:(2)]$

$=2$ × $(k+6)=4k+2$

$=2k+12=4k+2\\\\=4k+2=2k+12$

$=4k-2k=12-2\\\\=2k=10\\\\=k=\frac{10}{2}$

$k=5$

∴ The value of k is 5.