# Question #52ae1

Dec 31, 2014

$k = \pm \frac{5}{2}$

You have $4 {x}^{2} - 8 k x + 9 = 0$
and you know that the two solutions, ${x}_{2} \mathmr{and} {x}_{1}$, are related as:
${x}_{2} - {x}_{1} = 4$

You solve your equation with $k$ in place:
${x}_{1 , 2} = \frac{8 k \pm \sqrt{64 {k}^{2} - 4 \left(4 \cdot 9\right)}}{8}$
Giving:
${x}_{1 , 2} = \frac{8 k \pm \sqrt{64 {k}^{2} - 144}}{8}$
These are two solutions: one with the + sign and the other with the - sign. Substituting in: ${x}_{2} - {x}_{1} = 4$ you get
$\frac{8 k + \sqrt{64 {k}^{2} - 144}}{8} - \frac{8 k - \sqrt{64 {k}^{2} - 144}}{8} = 4$
rearranging and taking as common denominator $8$ you'll get:
$2 \cdot \left(\sqrt{64 {k}^{2} - 144}\right) = 32$
$\left(\sqrt{64 {k}^{2} - 144}\right) = 16$
$64 {k}^{2} - 144 = 256$
$k = \sqrt{\frac{400}{64}} = \pm \frac{5}{2}$