Question

What is the set of values of $"a"(a\ne 0)$ for which both zeros of the quadratic polynomials $f\left(x\right)=a{x}^2+(a-3)x+1$ lies only on one side of the Y-axis.

• A. $(-\infty ,1]\cup [9,\infty )$
• B. None of the above
• C. $[9,\infty )$
• D. $(0,1]\cup [9,\infty )$

Answer 1

The correct option is D $(0,1]\cup [9,\infty )$

Given: $f\left(x\right)=a{x}^2+(a-3)x+1,a\ne 0$
both zeros of the quadratic polynomials lies only on one side of the Y-axis.

To find: Set of values of $a$

Step-1: Use $D\ge 0$ and solve for $a$
Step-2: Also solve for $a$ when (i) both roots are negative (ii) Both roots are positive.
Step-3: Find the final set of values of a.

$D\ge 0$
$\Rightarrow (a-3{)}^2-4.a.1\ge 0$
$\Rightarrow (a-1)(a-9)\ge 0$
$\Rightarrow a\le 1$ or $a\ge 9$
$\Rightarrow a\in (-\infty ,1]$ or $a\in [9,\infty )$
$\Rightarrow a\in (-\infty ,1]\cup [9,\infty )$ but $a\ne 0$
$\therefore a\in (-\infty ,1]\cup [9,\infty )-\left\{0\right\}=A\left(Say\right)$


Step-2: case (i) Both roots are negative.

$-\frac{a-3}{a}<0$ and $\frac{1}{a}>0$
$\Rightarrow \frac{a-3}{a}>0$ and $a>0$
$\Rightarrow a>0$ and $a-3>0$
$\Rightarrow a>3$
$\Rightarrow a\in (3,\infty )=B$

Case: (ii) Both roots are positive.

$-\frac{a-3}{a}>0$ and $\frac{1}{a}>0$
$\Rightarrow \frac{a-3}{a}<0$ and $a>0$
$\Rightarrow 0<a<3$
$\therefore a\in (0,3)=C$

Range of $a,$ is $A\cap (B\cup C)=(0,1]\cup [9,\infty )$