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#