Question

If the roots of the equation \[{{x}^{2}}+2ax+b=0\] are real and distinct and they differ by at most 2m, then b lies in the interval
\[\begin{align}
  & 1)\left( {{a}^{2}},{{a}^{2}}+{{m}^{2}} \right) \\
 & 2)\left( {{a}^{2}}-{{m}^{2}},{{a}^{2}} \right) \\
 & 3)\left[ {{a}^{2}}-{{m}^{2}},{{a}^{2}} \right) \\
 & 4)None \\
\end{align}\]

Answer 1

Hint: It is important to know the general form a quadratic equation ,sum of roots and product of roots of the quadratic equation
General form : \[p{{x}^{2}}+qx+r=0\]
Sum of roots: \[\dfrac{-q}{p}\]
Product of roots: \[\dfrac{r}{p}\]
As the roots are real and distinct the discrimnant \[{{q}^{2}}-4pr>0\]. As mentioned in the question the difference between the roots should be found and the difference is less than or equal to 2m.

Complete step-by-step answer:
Now comparing the given equation\[{{x}^{2}}+2ax+b=0\] with the general form of a quadratic equation \[p{{x}^{2}}+qx+r=0\]
\[\begin{align}
  & p=1 \\
 & q=2a \\
 & r=b \\
\end{align}\]
Let the roots of the given quadratic equation be \[\alpha \] and \[\beta \]
Sum of roots = \[\alpha +\beta \] = \[\dfrac{-2a}{1}=-2a\]
Product of roots=\[\alpha \beta \] =\[\dfrac{b}{1}=b\]
It is now mentioned in the question that the difference between the roots is at most 2m
Which implies that
\[|\alpha -\beta |\le 2m\]
 By squaring on both sides of the above equation we get
\[\begin{align}
  & {{\left( \alpha -\beta \right)}^{2}}\le {{\left( 2m \right)}^{2}} \\
 & \\
\end{align}\]
As we know that
\[{{\left( \alpha -\beta \right)}^{2}}={{\left( \alpha +\beta \right)}^{2}}-4\alpha \beta \]
Substituting it in the above equation gives us
 \[{{\left( \alpha +\beta \right)}^{2}}-4\alpha \beta \le {{\left( 2m \right)}^{2}}\]
We know that \[\alpha +\beta =-2a,\alpha \beta =b\] substituting these values in the above equation gives
\[\begin{align}
  &\Rightarrow {{\left( -2a \right)}^{2}}-4\left( b \right)\le {{\left( 2m \right)}^{2}} \\
 &\Rightarrow 4{{a}^{2}}-4b\le 4{{m}^{2}} \\
 &\Rightarrow {{a}^{2}}-b\le {{m}^{2}} \\
 &\Rightarrow {{a}^{2}}-{{m}^{2}}\le b \\
 &\Rightarrow b\ge {{a}^{2}}-{{m}^{2}} \\
\end{align}\]
As the roots are real and distinct we also know that the discriminant should be greater than 0
\[{{q}^{2}}-4pr>0\] substituting the values of q, p and r we get
\[\begin{align}
  & {{\left( 2a \right)}^{2}}-4\left( 1 \right)\left( b \right)>0 \\
 &\Rightarrow 4{{a}^{2}}-4b>0 \\
 &\Rightarrow {{a}^{2}}-b>0 \\
 &\Rightarrow {{a}^{2}}>b \\
 &\Rightarrow b<{{a}^{2}} \\
\end{align}\]
So here we have
\[\begin{align}
  & {{a}^{2}}-{{m}^{2}}\le b<{{a}^{2}} \\
 &\Rightarrow b\in \left[ {{a}^{2}}-{{m}^{2}},{{a}^{2}} \right) \\
\end{align}\]
Hence the answer is Option 3

So, the correct answer is “Option 3)”.

Note: While solving questions of this type it is important to take care of substitutions and the types of brackets in the option whether it is a closed or an open bracket. Student can also solve this problem by directly considering the roots of a quadratic equation which are
     \[\dfrac{-q+\sqrt{{{q}^{2}}-4pr}}{2p},\dfrac{-q-\sqrt{{{q}^{2}}-4pr}}{2p}\]
Instead of using \[\alpha \] and\[\beta \] by substituting the values and finding the difference of roots.