# 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"

May 5, 2018

The quadratic formula is:

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

The first root is:

${x}_{1} = \frac{- b - \sqrt{{b}^{2} - 4 a c}}{2 a} \text{ [1]}$

The second root is:

${x}_{2} = \frac{- b + \sqrt{{b}^{2} - 4 a c}}{2 a} \text{ [2]}$

We want:

${x}_{2} - {x}_{1} = \sqrt{69} \text{ [3]}$

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

$\frac{- b + \sqrt{{b}^{2} - 4 a c}}{2 a} - \frac{- b - \sqrt{{b}^{2} - 4 a c}}{2 a} = \sqrt{69}$

Combine the two fractions over the common denominator:

$\frac{- b + \sqrt{{b}^{2} - 4 a c} + b + \sqrt{{b}^{2} - 4 a c}}{2 a} = \sqrt{69}$

Simplify:

$\frac{2 \sqrt{{b}^{2} - 4 a c}}{2 a} = \sqrt{69}$

Substitute $a = 1 , b = k , \mathmr{and} c = 3$:

$\frac{2 \sqrt{{k}^{2} - 4 \left(1\right) \left(3\right)}}{2 \left(1\right)} = \sqrt{69}$:

Multiply $- 4 \left(1\right) \left(3\right) = - 12 \mathmr{and} 2 \left(1\right) = 2$:

$\frac{2 \sqrt{{k}^{2} - 12}}{2} = \sqrt{69}$

$\frac{2}{2}$ cancels:

$\sqrt{{k}^{2} - 12} = \sqrt{69}$

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 +-):

$k = \pm 9$