Question

if the sum of the zeros of the quadratic polynomial k y square - 2 Y - 3 K is equal to twice their product
Find the value of k.​

Answer 1

Answer :

k = -⅓

Note:

★ The possible values of the variable for which the polynomial becomes zero are called its zeros .

★ A quadratic polynomial can have atmost two zeros .

★ The general form of a quadratic polynomial is given as ; ax² + bx + c .

★ If α and ß are the zeros of the quadratic polynomial ax² + bx + c , then ;

• Sum of zeros , (α + ß) = -b/a

• Product of zeros , (αß) = c/a

Solution :

Here ,

The given quadratic polynomial is ;

ky² - 2y - 3k .

Now ,

Comparing the given quadratic polynomial with the general quadratic polynomial ay² + by + c ,

We have ;

a = k

b = -2

c = -3k

Now ,

=> Sum of zeros = -b/a

=> Sum of zeros = -(-2)/k

=> Sum of zeros = 2/k

Also ,

=> Product of zeros = c/a

=> Product of zeros = -3k/k

=> Product of zeros = -3

Now ,

According to the question ,

=> Sum of zeros = 2 × Product of zeros

=> 2/k = 2 × (-3)

=> 1/k = -3

=> k = -⅓

Hence , k = -⅓

Answer 2

Answer:

• Given Polynomial : ky² - 2y - 3k
• a = k
• b = - 2
• c = - 3k

$\underline{\bigstar\:\textsf{According to the given Question :}}$

$:\implies\textsf{Sum of Zeroes = Twice of Product of Zeroes}\\\\\\:\implies\sf \dfrac{-\:b}{a}=2 \times \dfrac{c}{a}\\\\\\:\implies\sf - \:b =2c\\\\\\:\implies\sf -( - 2) =2 \times ( - 3k)\\\\\\:\implies\sf 2 = 2 \times - 3k\\\\\\:\implies\sf \dfrac{2}{2} = -3k\\\\\\:\implies\sf1 = - 3k\\\\\\:\implies\sf \dfrac{1}{ - 3} = k\\\\\\:\implies\underline{\boxed{\sf k = \dfrac{ - \:1}{3} }}$