Question

The number a is real and such that the equation x^2 +2 (a - 1) x - a + 7 = 0 has

  two different real negative solutions. One can then conclude that

  (a) a <−2; (b) 3 <a <7; (c) it is impossible; (d) none of (a) - (c).

Answer 1

Given equation-

$x^2+2(a-1)x-a+7=0$

Let the required roots be -m and -n

then sum of roots $-m-n=-2(a-1)~~~~-(1)$

Also Product of roots $-m\times -n=-a+7~~~~~-(2)$

$m+n=2a-2 \text{ and } mn=7-a$

As, $mn>0\Rightarrow 7-a>0\Rightarrow a<7$

Also, $m+n>0\Rightarrow 2a-2>0\Rightarrow a>1$

So $1<a<7.$