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4-2.Quadratic Equations and Inequations
easy
If the roots of ${x^2} + x + a = 0$exceed $a$, then
A
$2 < a < 3$
B
$a > 3$
C
$ - 3 < a < 3$
D
$a < - 2$
Solution
(d) If the roots of the quadratic equation $a{x^2} + bx + c = 0$ exceed a number k, then $a{k^2} + bk + c > 0$ if $a > 0,$ ${b^2} – 4ac \ge 0$ and sum of the roots $ > 2k$
Therefore, if the roots of ${x^2} + x + a = 0$ exceed a number a, then
${a^2} + a + a > 0,1 – 4a \ge 0$ and $ – 1 > 2a$
==> $a(a + 2) > 0,$$a \le \frac{1}{4}$and $a < – \frac{1}{2}$
==> $a > 0\,{\rm{or}}\,a < – 2,a < \frac{1}{4}$and $a < – \frac{1}{2}$
Hence $a < – 2$.
Standard 11
Mathematics
