Question

If $\alpha $and $\beta \left( \ne 0 \right)$ are the roots of the quadratic equation ${{x}^{2}}+\alpha x-\beta =0$ , then the quadratic expression $-{{x}^{2}}+\alpha x+\beta $, where $x\in \mathbb{R}$ has
 (a) Least value $-\dfrac{1}{4}$
(b) Least value $-\dfrac{9}{4}$
(c) Greatest value $\dfrac{1}{4}$
(d) Greatest value $\dfrac{9}{4}$

Answer 1

Hint: We have, in the problem, we are given two roots of a quadratic equation ${{x}^{2}}+\alpha x-\beta =0$. So, we will get with it by using the properties of quadratic equations. Thus we will get a relation with the roots and then simplify that we will get the value of the roots. Using the values and simplifying the next equation we will get the greatest value or least value of the problem.

Complete step by step solution:
According to the problem, $\alpha $and $\beta \left( \ne 0 \right)$ are the roots of the quadratic equation ${{x}^{2}}+\alpha x-\beta =0$.
And we also know from the properties of quadratic equations, if y and z be two roots of a quadratic equation $a{{x}^{2}}+bx+c=0$ , then, $y+z=-\dfrac{b}{a}$ and $yz=\dfrac{c}{a}$ ,
Thus, we can conclude, $\alpha +\beta =-\alpha $ and $\alpha \beta =-\beta $ ,
Now, from, $\alpha \beta =-\beta $,
We get, $\alpha \beta +\beta =0$
$\Rightarrow \beta \left( \alpha +1 \right)=0$
As, $\beta \left( \ne 0 \right)$, we can say, $\alpha =-1$
Substituting the value in, $\alpha +\beta =-\alpha $, we are getting, $\beta =-2.\left( -1 \right)=2$
Now, putting the values in $-{{x}^{2}}+\alpha x+\beta $we have, $-{{x}^{2}}-x+2$
Again, $-{{x}^{2}}-x+2$ can be written as,
$\Rightarrow -\left[ {{x}^{2}}+x-2 \right]$
$\Rightarrow -\left[ {{x}^{2}}+x+\dfrac{1}{4}-\dfrac{9}{4} \right]$
To try using algebraic properties,
$\Rightarrow -\left[ {{x}^{2}}+2.\dfrac{1}{2}.x+{{\left( \dfrac{1}{2} \right)}^{2}}-\dfrac{9}{4} \right]$
$\Rightarrow -\left[ {{\left( x+\dfrac{1}{2} \right)}^{2}}-\dfrac{9}{4} \right]$
Multiplying the negative sign inside, we are having,
$\Rightarrow \dfrac{9}{4}-{{\left( x+\dfrac{1}{2} \right)}^{2}}$
Now, the term ${{\left( x+\dfrac{1}{2} \right)}^{2}}$ is always positive as square of a number is always positive be it negative or positive.
So, the value of the term $-{{x}^{2}}-x+2$ will always be less than $\dfrac{9}{4}$ as the term is getting subtracted by ${{\left( x+\dfrac{1}{2} \right)}^{2}}$which is always greater than zero.
Then, we can easily say, the maximum or the greatest value of the term $-{{x}^{2}}-x+2$will be $\dfrac{9}{4}$.

So, the correct answer is “Option d”.

Note: In this problem we have used the properties of quadratic equations, saying, if y and z be two roots of a quadratic equation $a{{x}^{2}}+bx+c=0$ , then, $y+z=-\dfrac{b}{a}$ and $yz=\dfrac{c}{a}$. This condition is only valid under the condition $a\ne 0$, as otherwise, by dividing terms with ‘a’ we will get indefinite solutions.