Question

find the value of k so that the zeroes of the quadratic polynomial 3x2-kx+14 are in the ratio 7:6

Answer 1

let the zeros be 7y and 6y.
alpha + beta = -b/a
Alpha+beta= k/3.
7y+6y= k/3
13y= k/3
y= k/39. ........ (1)
Alpha×beta = c/a
7y×6y= 14/3
= 42y2= 14/3
= y2= 14/3×1/42
= y2= 1/9
= y= 1/3
subsitute the value of y in (1)
1/3= k/39
= k= 13.

Answer 2

Answer:

k = 13

k = 13Step-by-step explanation:

Sol:

 If α,β are the roots of the equation ax2 + bx + c = 0.

 α + β = -b / a and αβ = c / a.

 If α,β are in the ratio 7:6

then

α / β = 7 / 6.

 ∴ α  = 7β / 6.

 If α,β are the roots of the equation 3x2 - kx + 14 = 0. 

α + β = k / 3 ----------(1) 

And αβ = 14 / 3

⇒ 7β^2 / 6 = 14 / 3 .           [ α  = 7β / 6] 

⇒ β^2  = 4 .

 ∴ β = 2 and α  = 7 / 3. 

Substitute α , β are roots in equation (1) 

⇒ (7 / 3) + 2 = k / 3

 ⇒ 13 / 3 = k / 3

 ∴ k = 13.

Hope it helps.

:)