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"

1 Answer
May 5, 2018

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