Question

If the equation $x^2+2(a+1)x+9a-5=0$ has only negative root, then

• A. $a>0$
• B. $a\ge 0$
• C. $a\le -0$
• D. $a\ge 6$

Answer 1

The correct option is D $a\ge 6$

Condition for negative root is only that,

$D\ge 0$

Sum of roots < 0

Product of roots > 0

So here for $x^2+2(a+1)x+9a-5=0$

$4(a+1{)}^2-4\times (9a-5)\ge 0$

$4a^2+4+8a-36+20a\ge 0$

$a^2-7a+6\ge 0$

$(a-6)(a-1)\ge 0$

$a\in (-\infty ,1]\cup [6,\infty )$ .....(1)

Second condition,

$-2(a+1)<0$

$a+1>0$

$a\in (-1,\infty )$ .......(2)

And third condition,

$9a-5>0$

$a\in (\frac{5}{9},\infty )$ .....(3)

Now, intersection of 1, 2, 3 will be the final answer, so, we get,

$a\in [6,\infty )$

Thus, option D is correct.