Question

if the equation x^2+2(k+1)x+9k-5= 0 has only negative roots, k must be

Answer 1

Discriminant <0
4(k+1)² -4(9k-5) <0
k² + 1 +2k -9k +5<0
k² -7k +6<0
(k-1)(k-6)<0
k € (1,6)

Answer 2

Answer:

The condition for k is k must be greater than $\frac{5}{9}$              

Step-by-step explanation:

Given : Equation $x^2+2(k+1)x+9k-5= 0$ has only negative roots.

To find : The value of k?

Solution :

Equation $x^2+2(k+1)x+9k-5= 0$ has only negative roots.

When the quadratic equation has only negative roots then their product of zeros is always positive.

Product of zeros - $\alpha \beta=\frac{c}{a}$

Where, c=9k-5 and a=1

$\alpha \beta=\frac{9k-5}{1}$

$\alpha \beta>0$

$9k-5>0$

$9k>5$

$k>\frac{5}{9}$

Therefore, The condition for k is k must be greater than $\frac{5}{9}$