Question

The given equation is \[\left( l+{{m}^{2}} \right){{x}^{2}}-2\left( 1+3m \right)x+\left( 1+8m \right)=0\] . The number of values of m for which the given equation has no real solution is
$\begin{align}
  & \text{a) 3} \\
 & \text{b) }\infty \\
 & \text{c) 2} \\
 & \text{d) 1} \\
\end{align}$

Answer 1

Hint: Now we are given with equation \[\left( l+{{m}^{2}} \right){{x}^{2}}-2\left( 1+3m \right)x+\left( 1+8m \right)=0\] . We know that for any equation $a{{x}^{2}}+bx+c=0$ the condition for non-real solution is ${{b}^{2}}-4ac<0$. Hence we will use this equation to find conditions on x.

Complete step by step answer:
Now we are given with the equation \[\left( l+{{m}^{2}} \right){{x}^{2}}-2\left( 1+3m \right)x+\left( 1+8m \right)=0\]
Now the equation \[\left( l+{{m}^{2}} \right){{x}^{2}}-2\left( 1+3m \right)x+\left( 1+8m \right)=0\] is in the form of $a{{x}^{2}}+bx+c=0$ .
Any equation of the form $a{{x}^{2}}+bx+c=0$ is called a quadratic equation.
Now we know that for a quadratic equation we have Discriminant D as $D={{b}^{2}}-4ac$ .
Now the nature of roots depends upon the value of the discriminant.
If we have D = 0, then the roots are real and equal.
If we have D > 0, then we have roots are real and distinct.
And finally, if we have D < 0, then there are no real roots.
Now since we need to find condition on m such that there are no real roots we will use D < 0 in equation \[\left( l+{{m}^{2}} \right){{x}^{2}}-2\left( 1+3m \right)x+\left( 1+8m \right)=0\] .
Now for the equation \[\left( l+{{m}^{2}} \right){{x}^{2}}-2\left( 1+3m \right)x+\left( 1+8m \right)=0\] we have
$D={{\left( 2\left( 1+3m \right) \right)}^{2}}-4\left( 1+{{m}^{2}} \right)\left( 1+8m \right)$
Now let us take D < 0. And solve the equation.
$\begin{align}
  & {{\left( 2\left( 1+3m \right) \right)}^{2}}-4\left( 1+{{m}^{2}} \right)\left( 1+8m \right)<0 \\
 & \Rightarrow 4{{\left( 1+3m \right)}^{2}}-4\left( 1+{{m}^{2}} \right)\left( 1+8m \right)<0 \\
\end{align}$
Now dividing the whole equation by 4 we get.
${{\left( 1+3m \right)}^{2}}-\left( 1+{{m}^{2}} \right)\left( 1+8m \right)<0$
Now using the formula ${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$ we get.
$\begin{align}
  & \left( 1+6m+9{{m}^{2}} \right)-\left( 1+8m+{{m}^{2}}+8{{m}^{3}} \right)<0 \\
 & \Rightarrow \left( 1+6m+9{{m}^{2}}-1-8m-{{m}^{2}}-8{{m}^{3}} \right)<0 \\
 & \Rightarrow \left( -8{{m}^{3}}+8{{m}^{2}}-2m \right)<0 \\
\end{align}$
Now taking – 2 common we get.
$\begin{align}
  & -2m\left( 4{{m}^{2}}-4m+1 \right)<0 \\
 & \Rightarrow -2m\left( 4{{m}^{2}}-2m-2m+1 \right)<0 \\
 & \Rightarrow -2m\left( 2m\left( 2m-1 \right)-1\left( 2m-1 \right) \right)<0 \\
 & \Rightarrow -2m{{\left( 2m-1 \right)}^{2}}<0 \\
\end{align}$
Now we need to find out for how many values of m the equation $-2n{{\left( 2m-1 \right)}^{2}}<0$ is true.
Now we know that ${{\left( 2m-1 \right)}^{2}}$ is always positive.
And we know that $\left( - \right)\times \left( + \right)=\left( - \right)$
Hence we need – 2 m to always be negative. And hence m > 0.
Hence for all values of m such that m > 0, we have $-2m\left( 4{{m}^{2}}-4m+1 \right)<0$
Hence have the have $m\in \left( 0,\infty \right)$ . Now since there are infinite numbers between 0 to infinity
We have infinite possible m.
Option b is the correct option.

Note: Note not to get confused between the three conditions of Discriminant if we have ${{b}^{2}}-4ac$ positive then the roots are real and distinct. If we have ${{b}^{2}}-4ac$equal to zero then the roots are repeating and if we have ${{b}^{2}}-4ac$ as a negative value then there are no real solutions of the quadratic.