Question

3. If the sum of the zeroes of the quadratic
polynomial ky^2+2y-3k is equal to twice their
product, find value of k.​

Answer 1

Let @ and ß to be the zeros of above mentioned quadratic polynomial.

Consider p(y) to be the polynomial.

Now,

p(y)=ky²+2y-3k

Given condition:

Sum of zeros=2×product of zeros

Here,

Sum of zeros: -x coefficient/x² coefficient

→@+ß= -2/k

Product of zeros: constant term/x² coefficient

→@ß= -3k/k= -3

According to given condition,

@+ß=2@ß

→ -2/k= -6

→k= 2/6

→k=1/3

For k=1/3,the above polynomial has zeros such that their sum is twice their product

Answer 2

Answer:

$\large{\underline{\boxed{\sf k = \dfrac{1}{3}}}}$

Step-by-step explanation:

Let the zeroes of the quadratic polynomial be α and β.

F(y) = ky² + 2y - 3k

On comparing the given equation with ay² + by + c,

• a = k
• b = 2
• c = - 3k

Sum of zeroes = -b/a

⇒ α + β = - 2/k

Product of zeroes = c/a

⇒ αβ = -3k/k

⇒ αβ = - 3

Now, it is given that ;

α + β = 2αβ

$\implies \sf - \frac{2}{k} = 2( -3 ) \\ \\ \implies \sf - \frac{2}{k} = - 6 \\ \\ \implies \sf 6k = 2 \\ \\ \implies \sf k = \frac{2}{6} \\ \\ \boxed{ \bf \therefore \: k = \frac{1}{3}}$

Hence, the value of k is $\bf \dfrac{1}{3}$ .