Question

If the sum of the square of zeroes of the polynomial 6 x^2 + x + k is 25/36 , find the value of k.

Answer 1

We have :-

P(x) =6x²+x+k

Sum of square if zeroes is 25/36

Let the zeroes of p(x) be a and b

We already know that :-

Sum of zeroes of a polynomial = - b/a

Product of zeroes = c/a

Now,

${a}^{2} + {b}^{2} = \frac{25}{36} \\ \\ \\ {(a + b)}^{2} - 2ab = \frac{25}{36}$

Now, From the polynomial we have values of :-

$a + b = \frac{ - 1}{6} \\ \\ \\ab = \frac{k}{6}$

Now, we have the following :-

${( \frac{ - 1}{6})}^{2} - \frac{2k}{6} = \frac{25}{36} \\ \\ \\ \frac{1}{36} - \frac{12k}{36} = \frac{25}{36} \\ \\ \\1 - 12k = 25 \\ \\ \\12k = ( - 24) \\ \\ \\k = \frac{( - 24)}{12} \\ \\ \\k = ( - 2)$

Answer 2

Answer:

Step-by-step explanation:

Let α, β are two zeros of given Quadratic equation.

comparing the given eqn. with ax^2 + bx + c

a=6   b= 1  and c = k

then α + β = - b / a = - 1 / 6  

αβ = c / a = k / 6  

α² + β² = 25 / 36    ( given )

(α + β)² - 2αβ = 25 / 36  

( -1 / 6 )² - 2 ( k / 6 ) = 25 / 36  

( 1 / 36 ) -  ( 2k / 6 ) = 25 / 36

( 1 / 36 ) - ( k / 3 ) = 25 / 36  

( 1 / 36 ) - ( 12k / 36 ) = 25 / 36  

( 1 - 12k ) / 36  = 25  / 36

36 ( 1 - 12k ) / 36  = 25

1 - 12k = 25

-12k = 25 - 1 = 24  

k = 24 / -12

k = -2