Question

If the equation \[a{{x}^{2}}+bx+6=0\] has real roots where \[a,b\in R\], then greatest value of \[3a+b\] is
(a) 4
(b) -1
(c) -2
(d) 1

Answer 1

Hint: For solving this question you should know about the distinct and real roots of any quadratic equation. If that equation contains only real roots then there is no distinct root and thus the value of \[{{b}^{2}}-4ac\] is always zero here. So, find the \[3a+b\] here.


Complete step-by-step solution:
According to the question it is asked to find the value \[3a+b\] if the quadratic equation \[a{{x}^{2}}+bx+6=0\] contains real roots here.
So, as we know, the equation does not have two distinct real roots. So, the value of \[{{b}^{2}}-4ac\] here will be equal to zero. And by applying this condition here we get,
\[\begin{align}
  & {{b}^{2}}-4ac=0 \\
 & {{b}^{2}}=4ac \\
\end{align}\]
Here, \[{{b}^{2}}=4\times a\times 6=24a\]
           \[{{b}^{2}}=24a\]
So, \[3a=\dfrac{{{b}^{2}}}{8}\]
 And \[3a+b=\dfrac{{{b}^{2}}}{8}+b=\dfrac{{{b}^{2}}+8b}{8}\]
                      \[=\dfrac{{{\left( b+4 \right)}^{2}}-16}{8b}\]
So, least value of \[{{\left( b+4 \right)}^{2}}=0\]
Hence, the least value of \[\dfrac{{{\left( b+4 \right)}^{2}}-16}{8b}=\dfrac{0-16}{8}=-2\]
So, the value of b will be ‘-2’.
So, the correct option is ‘(c)’.

Note: While solving these types of questions you have to mind that the roots of any quadratic equation can be solved by the \[{{b}^{2}}-4ac\]. And if the question asks for the values of any variable in this, then we will solve it by the help of this formula and find the value.