Question

For any quadratic expression $y={\text{x}}^2-b\text{x}+c,$ if the sum of it's roots is equal to the product of it's roots and $y\ge -8\forall x\in R.$ Then find the range of $b.$

• A. $[-8,16]$
• B. $[-4,8]$
• C. $[-8,4]$
• D. $[-8,8]$

Answer 1

The correct option is B $[-4,8]$

Given : $\text{y}={\text{x}}^2-b\text{x}+c\ge -8\forall \text{x}\in R\cdots \left(i\right)$
For ${\text{x}}^2-b\text{x}+c=0,$
Sum of roots $=$ Product of roots
By comparing with the standard quadratic equation $\text{y}=p{\text{x}}^2+q\text{x}+r,$ we get:
$p=1,q=-b,r=c$.
Here, $p$ is positive which means it is an upward opening parabola and we know the vertex of upward opening parabola i.e. $(-\frac{q}{2p},-\frac{D}{4p})$, therefore the range of the quadratic equation is $[-\frac{D}{4p},\infty )$.

Therefore the minimum value of the equation is at $\text{x}=-\frac{q}{2p}=-\frac{-b}{2.1}=\frac{b}{2}$
$\therefore \text{y}=(\frac{b}{2}{)}^2-b\times \frac{b}{2}+c=-\frac{b^2}{4}+c$
$\Rightarrow \text{y}=-\frac{b^2}{4}+c\ge -8\cdot \cdot \left(ii\right)\left[\text{From}\right(i\left)\right]$

Also from the relation sum of roots = product of roots
$\Rightarrow -(-b)=c$
$\Rightarrow b=c\left(iii\right)$

$\Rightarrow -\frac{b^2}{4}+b\ge -8\left[\text{From}\right(ii\left)\text{and}\right(iii\left)\right]$

$\Rightarrow -b^2+4b\ge -32$
$\Rightarrow b^2-4b\le 32$
Add $4$ on both the sides to simplify the equation
$\Rightarrow b^2-4b+4\le 32+4$
$\Rightarrow (b-2{)}^2\le (6{)}^2$
$\Rightarrow -6\le (b-2)\le 6$
$\Rightarrow -4\le b\le 8$
$\Rightarrow b\in [-4,8]$