Question

Consider the equation $x^2+x+a-9<0$. The values of the real parameter $a$ so that the given equation has at least one positive solution are

• A. $(-\infty ,\frac{37}{4})$
• B. $(-\infty ,\infty )$
• C. $(3,\infty )$
• D. $(-\infty ,9)$

Answer 1

The correct option is D $(-\infty ,9)$

Given,
$x^2+x+a-9<0$
$(x+\frac{1}{2}{)}^2+a-\frac{37}{4}<0$
has atleast one positive solution
i.e.,atleast one solution is such that $x>0$
$x+\frac{1}{2}>\frac{1}{2}$
$(x+\frac{1}{2}{)}^2>\frac{1}{4}$
$(x+\frac{1}{2}{)}^2-\frac{1}{4}>0$
=>$(x+\frac{1}{2}{)}^2+a-\frac{37}{4}<0<(x+\frac{1}{2}{)}^2-\frac{1}{4}$
=>$(x+\frac{1}{2}{)}^2+a-\frac{37}{4}<(x+\frac{1}{2}{)}^2-\frac{1}{4}$
$a<9$
=>$a\in (-\infty ,9)$
Hence, option D is correct.