Question

If the roots of the equation \[{\lambda ^2} + 8\lambda + {\mu ^2} + 6\mu = 0\] are real, then \[\mu \] lies between
A. \[ - 2\] and \[8\]
B. \[ - 3\] and \[6\]
C. \[ - 8\] and \[2\]
D. \[ - 6\] and \[3\]

Answer 1

Hint: The given equation is a quadratic equation in \[\lambda \] if \[\mu \] is supposed to be a constant or vice versa. The roots of a quadratic equation are real if and only if the discriminant of the equation is non-negative. Find the discriminant \[\left( D \right)\] and set an inequation \[D \ge 0\]. Solve the equation and obtain the required answer.

Formula Used:
Discriminant of a quadratic equation \[a{x^2} + bx + c = 0\] is given by \[D = {b^2} - 4ac\]
If the roots of the given equation are real, then \[D \ge 0\]

Complete step by step solution:
The given equation is \[{\lambda ^2} + 8\lambda + {\mu ^2} + 6\mu = 0\].
If \[\mu \] can be assumed as a constant then it is a quadratic equation in \[\lambda \].
Comparing it with the general form of a quadratic equation \[a{x^2} + bx + c = 0\], we get
\[a = 1,b = 8\] and \[c = {\mu ^2} + 6\mu \]
\[\therefore \]Discriminant of the equation is
\[D = {\left( 8 \right)^2} - 4\left( 1 \right)\left( {{\mu ^2} + 6\mu } \right)\]
\[ = 64 - 4{\mu ^2} - 24\mu \]
If the roots of the given equation are real, then
\[D \ge 0\]
\[ \Rightarrow 64 - 4{\mu ^2} - 24\mu \ge 0\]
\[ \Rightarrow - 4\left( {{\mu ^2} + 6\mu - 16} \right) \ge 0\]
\[ \Rightarrow {\mu ^2} + 6\mu - 16 \le 0\]
Let \[f\left( \mu \right) = \left( {{\mu ^2} + 6\mu - 16} \right)\]
Then \[f\left( \mu \right) \le 0\]
Let us solve this inequation to find the solution set of \[\mu \].
Factorize the quadratic polynomial \[f\left( \mu \right)\].
\[f\left( \mu \right) = {\mu ^2} + 6\mu - 16\]
\[ = {\mu ^2} + 8\mu - 2\mu - 16\]
\[ = \mu \left( {\mu + 8} \right) - 2\left( {\mu + 8} \right)\]
\[ = \left( {\mu + 8} \right)\left( {\mu - 2} \right)\]
Now, \[f\left( \mu \right) = 0\]
\[ \Rightarrow {\mu ^2} + 6\mu - 16 = 0\]
\[ \Rightarrow \left( {\mu + 8} \right)\left( {\mu - 2} \right) = 0\]
\[ \Rightarrow \left( {\mu + 8} \right) = 0\] or \[\left( {\mu - 2} \right) = 0\]
\[ \Rightarrow \mu = - 8\] or \[\mu = 2\]
\[\therefore \]The solution set of the equation \[f\left( \mu \right) = 0\] is \[\left\{ { - 8,2} \right\}\].
To solve the inequation \[f\left( \mu \right) < 0\], let us divide the real number line into three subintervals \[\left( { - \infty , - 8} \right),\left( { - 8,2} \right)\] and \[\left( {2,\infty } \right)\].
If we take \[\mu = - 10 \in \left( { - \infty , - 8} \right)\], then \[f\left( \mu \right) = {\left( { - 10} \right)^2} + 6\left( { - 10} \right) - 16 = 100 - 60 - 16 = 24 > 0\]
If we take \[\mu = 0 \in \left( { - 8,2} \right)\], then \[f\left( \mu \right) = {\left( 0 \right)^2} + 6\left( 0 \right) - 16 = 0 + 0 - 16 = - 16 < 0\]
If we take \[\mu = 10 \in \left( {2,\infty } \right)\], then \[f\left( \mu \right) = {\left( {10} \right)^2} + 6\left( {10} \right) - 16 = 100 + 60 - 16 = 144 > 0\]
\[\therefore \]The solution set of the inequation \[f\left( \mu \right) < 0\] is \[\left( { - 8,2} \right)\].
Thus the solution set of the inequation \[f\left( \mu \right) \le 0\] is \[\left[ { - 8,2} \right]\].
Hence option C is correct.

Note: In the given quadratic equation, two variables \[\lambda \] and \[\mu \] exist. So, the equation can be said quadratic either in \[\lambda \] or in \[\mu \] but we assumed that it is quadratic in \[\lambda \] because the question requires the domain of \[\mu \] so that the equation may have real roots.