Question

If
$if \: the \: sum \: of \: the \: polynomial \: f{t} = kt {}^{2} + 2t + 3k \: is \: equal \: to \: their \: product.find \: k$
​

Answer 1

Given :

$\bold{f(t) = kt^{2} + 2t + 3k}$

and

• Sum of zeroes = product of zeroes

To find:

• K = ?

Solution :

Here :

• a = k

• b= 2

• c= 3

=> sum of zeroes = product of zeroes

=> -b/a = c/a

=> -(2/k)= 3k/k

=> -2/k = 3

=> k = -(2/3)

Answer 2

GIVEN:-

f(t)= kt²+2t +3k

SOLUTION:-

➠ f(t)= kt² + 2t + 3k

here, ➠ • a = K, • b = 2 , • c = 3k

NOW,

➠ .°. sum of zeroes = -b/c

➠ .°. α + β = -2 /k

➠ product of zeroes = c/a

➠ .°. α × β = 3k / k

➠ .°. sum of zeroes = product of zeroes

➠ .°. α + β = α × β

➠ .°. ( -2 /k) = (3k / k)

➠ .°. -2/k = 3

➠.°. 3 = -2 / k

➠ .°. k = -2 /3

So, the value of K is -2/3.