Question

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

• A. $(-\infty ,9)$
• B. $(\frac{37}{4},\infty )$
• C. $(-\infty ,\frac{37}{4})$
• D. none of these

Answer 1

The correct option is C $(-\infty ,\frac{37}{4})$

Given,
$x^2+x+a-9<0$ has atleast one negative solution
=>$x<0$
$(x+\frac{1}{2}{)}^2+a-\frac{37}{4}<0$
$f\left(x\right)=(x+\frac{1}{2}{)}^2+a-\frac{37}{4}$

For $x<0,f\left(x\right)$ lies between $a-\frac{37}{4}$(at $x=-\frac{1}{2}$) and $\infty$ (when $x\approx \infty$)
=>$a-\frac{37}{4}$

So, $f\left(x\right)<0$ to have atleast one negative solution
=>$a-\frac{37}{4}<0$
$a<\frac{37}{4}$
=>$a\in (-\infty ,\frac{37}{4})$
Hence, option C is correct.