# If a > 0 and discriminant

Question:

If $a>0$ and discriminant of $a x^{2}+2 b x+c$ is negative, then

$\Delta=\left|\begin{array}{ccc}a & b & a x+b \\ b & c & b x+c \\ a x+b & b x+c & 0\end{array}\right|$ is

(a) positive

(b) $\left(a c-b^{2}\right)\left(a x^{2}+2 b x+c\right)$

(c) negative

(d) 0

Solution:

(c) negative

Discriminant D of $a x^{2}+2 b x+c=(2 b)^{2}-4 a c<0 \quad$ [Given]

$\Rightarrow 4 \mathrm{~b}^{2}-4 \mathrm{ac}<0$

$\Rightarrow \mathrm{b}^{2}-\mathrm{ac}<0$, where $\mathrm{a}>0 \quad \ldots(1)$

$\Delta=\mid \begin{array}{lll}\mathrm{a} & \mathrm{b} & \mathrm{ax}+\mathrm{b}\end{array}$

b $\quad$ c $\quad$ bx $+c$

$a x+b \quad b x+c \quad 0 \mid$

$=\mid \begin{array}{lll}a x & b x & a x^{2}+b x\end{array}$

$b \quad c \quad b x+c$

$\begin{array}{lll}\mathrm{ax}+\mathrm{b} & \mathrm{bx}+\mathrm{c} & 0\end{array}$     [Applying $\mathrm{R}_{1} \rightarrow \mathrm{x} \mathrm{R}_{1}$ ]

$=\frac{1}{x}\left|\mathrm{ax}+\mathrm{b} \quad \mathrm{bx}+\mathrm{c} \quad \mathrm{ax}^{2}+\mathrm{bx}+\mathrm{bx}+\mathrm{c} \quad \mathrm{b} \quad \mathrm{c} \quad \mathrm{bx}+\mathrm{cax}+\mathrm{b} \quad \mathrm{bx}+\mathrm{c} \quad 0\right|$

[Applying $\mathrm{R}_{1} \rightarrow \mathrm{R}_{1}+R_{2}$ ]

$=\frac{1}{x} \mid \begin{array}{lll}0 & 0 & a x^{2}+2 b x+c\end{array}$

b c $\quad b x+c$

$a x+b \quad b x+c \quad 0$            [Applying $\mathrm{R}_{1} \rightarrow \mathrm{R}_{1}-R_{3}$ ]

$=\frac{1}{x}\left\{\mathrm{ax}^{2}+2 \mathrm{bx}+\mathrm{c}\left|\begin{array}{cc}\mathrm{b} & \mathrm{c} \\ \mathrm{ax}+\mathrm{b} & \mathrm{bx}+\mathrm{c}\end{array}\right|\right\}$       [Expanding along $\mathrm{R}_{1}$ ]

$=\frac{1}{x}\left(a x^{2}+2 b x+c\right)\left(b^{2} x+b c-a c x-b c\right)$

$=\frac{1}{x}\left(a x^{2}+2 b x+c\right) x\left(b^{2}-a c\right)$

$=\left(a x^{2}+2 b x+c\right)\left(b^{2}-a c\right)<0$     [From eq. (1)]

$\Rightarrow \Delta<0$