If-f-x-x-2-8-x-a-8-1-has-2-diffrent-real-roots-in-7a-find-the-range-of-a-gt-0-

Question Number 185764 by CrispyXYZ last updated on 27/Jan/23

If f (x) = - \frac{x^{2}}{8} + x - \frac{a}{8} - 1 has 2 diffrent real

roots in (\sqrt{7 a}, + \infty), find the range of a > 0 .

Answered by CrispyXYZ last updated on 28/Jan/23

Let g (x) = f (x + \sqrt{7 a})

= - \frac{x^{2}}{8} + (1 - \frac{\sqrt{7 a}}{4}) x - a + \sqrt{7 a} - 1 (x > 0, a > 0)

So we get :

g (0) < 0 and Δ > 0 and symmetry axis = 4 - \sqrt{7 a} > 0

\Rightarrow a \in (0, \frac{5 - \sqrt{21}}{2})

Answered by Rajpurohith last updated on 27/Jan/23

Commented by mr W last updated on 28/Jan/23

w r o n g!

0 < a < \frac{5 - \sqrt{21}}{2}

Commented by Rajpurohith last updated on 21/Jun/23

y e s!! I m s o r r y i t s 0 < a < \frac{5 - \sqrt{21}}{2}