Quadratic Equation in two Variables - Practice Questions & MCQ

The general quadratic equation ax² + 2hxy + by² +2gx + 2fy + c = 0 can be resolved into linear factors 

$
a x^2+2 h x y+b y^2+2 g x+2 f y+c=0
$

we can write

$
a x^2+2(h y+g) x+\left(b y^2+2 f y+c\right)=0
$

equation (ii) is a quadratic form in terms of $x$
using the formula to get roots of quadratic equation

$
\begin{aligned}
& \therefore \mathrm{x}=\frac{-2(\mathrm{hy}+\mathrm{g}) \pm \sqrt{4(\mathrm{hy}+\mathrm{g})^2-4 \mathrm{a}\left(\mathrm{by}^2+2 \mathrm{fy}+\mathrm{c}\right)}}{2 \mathrm{a}} \\
& \Rightarrow \mathrm{x}=\frac{-(\mathrm{hy}+\mathrm{g}) \pm \sqrt{\mathrm{h}^2 \mathrm{y}^2+2 \mathrm{ghy}+\mathrm{g}^2-\mathrm{aby}^2-\mathrm{a} 2 \mathrm{fy}-\mathrm{ac}}}{\mathrm{a}} \\
& \Rightarrow \mathrm{ax}+\mathrm{hy}+\mathrm{g}= \pm \sqrt{\mathrm{h}^2 \mathrm{y}^2+2 \mathrm{ghy}+\mathrm{g}^2-\mathrm{aby}^2-\mathrm{a} 2 \mathrm{fy}-\mathrm{ac}} \\
& \Rightarrow a x+h y+g= \pm \sqrt{\left(h^2-a b\right) y^2+2(g h-a f) y+g^2-a c}
\end{aligned}
$
 

The expression (i) can be resolved into linear factors if 

$\left(h^2-a b\right) y^2+2(g h-a f) y+g^2-a c$ is a perfect square and $h^2-a b>0$.
The conditions for which $\left(h^2-\mathrm{ab}\right) \mathrm{y}^2+2(\mathrm{gh}-\mathrm{af}) \mathrm{y}+\mathrm{g}^2-\mathrm{ac}$ will be perfect square when

$
\begin{aligned}
& 4(\mathrm{gh}-\mathrm{af})^2-4\left(\mathrm{~h}^2-\mathrm{ab}\right)\left(\mathrm{g}^2-\mathrm{ac}\right)=0 \quad\left[\mathrm{D}=\mathrm{b}^2-4 \mathrm{ac}=0, \text { for perfact square }\right] \\
& \Rightarrow \mathrm{g}^2 \mathrm{~h}^2-2 \mathrm{ghaf}+\mathrm{a}^2 \mathrm{f}^2-\mathrm{h}^2 \mathrm{~g}^2+\mathrm{h}^2 \mathrm{ac}+\mathrm{abg}^2-\mathrm{a}^2 \mathrm{bc}=0 \\
& \Rightarrow \mathrm{abc}+2 \mathrm{fgh}-\mathrm{af}^2-\mathrm{bg}^2-\mathrm{ch}^2=0
\end{aligned}
$
 

Easy way to remember

$\begin{aligned} & \Delta=\mathrm{abc}+2 \mathrm{fgh}-\mathrm{af}^2-\mathrm{bg}^2-\mathrm{ch}^2=0 \\ & \text { i.e, }\left|\begin{array}{llll}a & h & g \\ h & b & f \\ g & f & c\end{array}\right|=0\end{aligned}$