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Answer 1 · 2017-03-19T18:39:36

$q = 6 and q = - 6$ $q = 6 and q = -6$

Let $r_{1} =$ $r_1 =$ the first root
Let $r_{2} =$ $r_2 =$ the second root
Given: $r_{1} - r_{2} = 2$ $r_1-r_2=2$

$(x - r_{1}) (x - r_{2}) = x^{2} - q x + 8$ $(x - r_1)(x - r_2) = x^2 - qx + 8$

$x^{2} - r_{2} x - r_{1} x + r_{1} r_{2} = x^{2} - q x + 8$ $x^2 -r_2x -r_1x + r_1r_2 = x^2-qx+8$

$- r_{1} - r_{2} = - q$ $-r_1-r_2 = -q$
$r_{1} - r_{2} = 2$ $r_1-r_2=2$
$r_{1} r_{2} = 8$ $r_1r_2=8$

Substitute for $r_{2} = \frac{8}{r_{1}}$ $r_2 = 8/r_1$ :

$- r_{1} - \frac{8}{r_{1}} = - q$ $-r_1-8/r_1 = -q$

$r_{1} - \frac{8}{r_{1}} = 2$ $r_1-8/r_1=2$

$q = r_{1} + \frac{8}{r_{1}} [1]$ $q = r_1+8/r_1" [1]"$

$r_{1}^{2} - 2 r_{1} - 8 = 0 [2]$ $r_1^2-2r_1-8=0" [2]"$

Factor equation [2]:

$(r_{1} - 4) (r_{1} + 2)$ $(r_1-4)(r_1+2)$

$r_{1} = 4 and r_{1} = - 2$ $r_1 = 4 and r_1 = -2$

$q = 4 + \frac{8}{4} and q = - 2 + \frac{8}{- 2}$ $q = 4 + 8/4 and q = -2+8/-2$

$q = 6 and q = - 6$ $q = 6 and q = -6$