Question

If the given quadratic equation ${{x}^{2}}+2\left( k+1 \right)x+9k-5=0$ has only negative roots. Find the interval in which k lies to satisfy this condition:
(a) $k\in \left( -\infty ,0 \right]$
(b) $k\in \left[ 0,\infty \right)$
(c) $k\in \left( \dfrac{5}{9},1 \right]\cup \left[ 6,\infty \right)$
(d) $k\in \left( -\infty ,6 \right]$

Answer 1

Hint: We have a quadratic equation ${{x}^{2}}+2\left( k+1 \right)x+9k-5=0$ which has only negative roots. We use properties of negative numbers for the sum of roots and product of roots. We use the fact that discriminant will be greater than or equal to zero for two real roots. Using all these we calculate the feasible interval for ‘k’.

Complete step by step answer:
Given that we have a quadratic equation ${{x}^{2}}+2\left( k+1 \right)x+9k-5=0$ which has only negative roots. We need to find the interval in k lies in order to satisfy the condition of having two negative roots.
Let us assume $\alpha $ and $\beta $ be the roots for the quadratic equation ${{x}^{2}}+2\left( k+1 \right)x+9k-5=0$.
We know that for a quadratic equation $a{{x}^{2}}+bx+c=0$. We have,
Sum of the roots = $\dfrac{-b}{a}$……(1)
Product of the roots = $\dfrac{c}{a}$……(2)
Applying equation (1) to quadratic equation ${{x}^{2}}+2\left( k+1 \right)x+9k-5=0$, we get
$\alpha +\beta =\dfrac{-2\left( k+1 \right)}{1}$.
$\alpha +\beta =-2\left( k+1 \right)$.
We know that sum of two negative numbers will also lead to a negative number. Since, $\alpha $ and $\beta $ are negative roots their sum will also be negative.
$\alpha +\beta <0......(3)$.
\[-2\left( k+1 \right)<0\].
$2\left( k+1 \right)>0$.
$\left( k+1 \right)>0$.
$k>-1$.
$k\in \left( -1,\infty \right)......(4)$.
The interval for ‘k’ which satisfies equation (3) is $\left( -1,\infty \right)$.
Applying equation (2) to quadratic equation ${{x}^{2}}+2\left( k+1 \right)x+9k-5=0$, we get
$\alpha .\beta =\dfrac{\left( 9k-5 \right)}{1}$.
$\alpha .\beta =(9k-5)$.
We know that multiplication of two negative numbers leads to a value of positive number. Since, $\alpha $ and $\beta $ are negative roots their product will be positive.
$\alpha .\beta >0......(5)$.
$\left( 9k-5 \right)>0$.
$9k>5$.
$k>\dfrac{5}{9}$.
$k\in \left( \dfrac{5}{9},\infty \right]......(6)$.
The interval for ‘k’ which satisfies equation (5) is $\left( \dfrac{5}{9},\infty \right)$.
We know that if any quadratic equation $a{{x}^{2}}+bx+c=0$ has two roots, then its discriminant $D={{b}^{2}}-4ac$ should be greater than or equal to zero $\left( D\ge 0 \right)$.
Using this condition for quadratic equation ${{x}^{2}}+2\left( k+1 \right)x+9k-5=0$, we get
${{\left( 2\left( k+1 \right) \right)}^{2}}-4.\left( 1 \right).\left( 9k-5 \right)\ge 0$.
$4{{\left( k+1 \right)}^{2}}-4.\left( 9k-5 \right)\ge 0$.
${{k}^{2}}+2k+1-9k+5\ge 0$.
${{k}^{2}}-7k+6\ge 0$.
${{k}^{2}}-6k-k+6\ge 0$.
$k\left( k-6 \right)-1\left( k-6 \right)\ge 0$.
$\left( k-1 \right).\left( k-6 \right)\ge 0......(7)$.
We know that the feasible interval for in-equality $\left( x-a \right).\left( x-b \right)\ge 0$and if $a < b$ is \[\left( x\le a \right)\cup \left( x\ge b \right)\]. Using this result for equation (7), we get
$\left( k\le 1 \right)\cup \left( k\ge 6 \right)$.
$k\in \left( -\infty ,1 \right]\cup \left[ 6,\infty \right)......(8)$.
We have got three different intervals for ‘k’ to satisfy the condition of negative roots. We know that if two or more intervals satisfy a condition, we estimate the common interval by taking intersection between all the intervals.
The required interval for ‘k’ to satisfy condition of negative roots for quadratic equation ${{x}^{2}}+2\left( k+1 \right)x+9k-5=0$ is \[\left( -1,\infty \right)\cap \left( \dfrac{5}{9},\infty \right)\cap \left[ \left( -\infty ,1 \right]\cup \left[ 6,\infty \right) \right]\].
We can observe that the common interval is $\left( \dfrac{5}{9},1 \right]\cup \left( 6,\infty \right]$.
∴ The feasible interval for ‘k’ is $k\in \left( \dfrac{5}{9},1 \right]\cup \left( 6,\infty \right]$.

So, the correct answer is “Option c”.

Note: We shouldn’t forget to check the case of discriminant $\left( D\ge 0 \right)$ as it may make changes in the final interval of ‘k’. We should not get confused with the concepts of inequalities in order to find the solution. We take intersection between the intervals as it gives us the common interval required.