Find the value of ‘k’ if one zero of p(x) = kx2
– 14 x + 8 is 6 times the other.

k = 3

Step-by-step explanation:

one zero of p(x) = kx²– 14 x + 8 is 6 times the other.

let the zeros are ⇒ a and 6a

the equation kx²– 14 x + 8

can be written as k ( x²/k– 14/k x + 8/k)

So, a * 6a = 8/k    ⇒ (1)

And a + 6a = 14/k ⇒ (2)

from (2)  k = 14/(a + 6a) = 14/(7a) = 2/a

∴ a = 2/k

Substitute at (1)

a * 6a = 8/k

2/k * 12/k = 8/k

solve for k

24/k = 8

k = 24/8 = 3

the value of k = 3

And the zeros will be 2/3 and 4

answer: integers are just whole numbers: positive, negative, and zero.

i = {-∞, -2, -1, 0, 1, 2, 3, ∞}

let x = 1st integer

let y = 2nd integer

x = 3y + 17   {equation 1}

xy = -24         {equation 2}

substitute 3y+17 in place of x n equation 2

and solve for y.

(3y+17)y = -24

3y2 + 17y = -24

3y2 + 17y + 24 = 0

from quadratic equation:

y = [-17±√(172-4(3)(24))]/(2(3))

y = (-17 ±√1)/6

y = (-17+1)/6   or     y = (-17-1)/6

y = -16/6           or     y = -18/6

y = -2 2/3         or   y = -3

since we are dealing with

y = -3

x = 3y+17 = -9+17 = 8

the integers are -3, 8

step-by-step explanation: