Question

The positive difference between the zeroes of the quadratic expression `x^2 + kx +3` is `sqrt(69)`. What are the possible values of k?

The book section is about the determinant, answer given by the book is : +/- 9.

I get that the determinant is: `k^2 -12`

and that 69 + 12 = 81, ans `sqrt(81) = 9)`. However, I am confuse by that direct math manipulation to put 12 in the square root. and the formulation use of "The difference between zeroes"

Answer 1

The quadratic formula is:

`x = (-b+-sqrt(b^2-4ac))/(2a)`

The first root is:

`x_1 = (-b-sqrt(b^2-4ac))/(2a)" [1]"`

The second root is:

`x_2 = (-b+sqrt(b^2-4ac))/(2a)" [2]"`

We want:

`x_2 - x_1 = sqrt69" [3]"`

Substitute equations [1] and [2] into equation [3]:

`(-b+sqrt(b^2-4ac))/(2a) - (-b-sqrt(b^2-4ac))/(2a) = sqrt69`

Combine the two fractions over the common denominator:

`(-b+sqrt(b^2-4ac) +b+sqrt(b^2-4ac))/(2a) = sqrt69`

Simplify:

`(2sqrt(b^2-4ac))/(2a) = sqrt69`

Substitute `a = 1, b = k, and c = 3`:

`(2sqrt(k^2-4(1)(3)))/(2(1)) = sqrt69`:

Multiply `-4(1)(3) = -12 and 2(1) = 2`:

`(2sqrt(k^2-12))/2 = sqrt69`

`2/2` cancels:

`sqrt(k^2-12) = sqrt69`

we can square both sides:

`k^2-12 = 69`

Add 12 to both sides:

`k^2 = 81`

Take the square root of both sides (don't forget the `+-)`:

`k = +-9`