1. Find the nature of the roots of the following quadratic equations. If the real roots exist, find them:
(i) $$2 x^2-3 x+5=0$$
(ii) $$3 x^2-4 \sqrt{3} x+4=0$$ [ JKBOSE – 2019, 15, 14, Most Imp. ]
(iii) $$2 x^2-6 x+3=0$$
Sol. (i) Consider equation $$2 x^2-3 x+5=0$$
Here, $$\quad a=2, b=-3, c=5$$
$$
\begin{aligned}
\quad \mathrm{D}=b^2-4 a c & =(-3)^2-4 \times 2 \times 5=9-40 \\
& =-31<0
\end{aligned}
$$
As $$\mathrm{D}<0$$, no real roots for the equation.
Sol. (ii) Consider equation $$3 x^2-4 \sqrt{3} x+4=0$$
Here, $$\quad a=3, b=-4 \sqrt{3}, c=4$$
$$
\begin{aligned}
\quad \mathrm{D}=b^2-4 a c & =(-4 \sqrt{3})^2-4 \times 3 \times 4 \\
& =48-48=0
\end{aligned}
$$
As $$\mathrm{D}=0$$, therefore, roots are real and equal
Roots are
$$
\begin{aligned}
\alpha & =\frac{-b+\sqrt{D}}{2 a}=\frac{-(-4 \sqrt{3})+0}{2 \times 3} \\
& =\frac{4 \sqrt{3}}{6}=\frac{2 \sqrt{3}}{3} \\
\beta & =\frac{-b-\sqrt{D}}{2 a}=\frac{-(-4 \sqrt{3})-0}{2 \times 3} \\
& =\frac{4 \sqrt{3}}{6}=\frac{2 \sqrt{3}}{3}
\end{aligned}
$$
and
Therefore, required roots are $$\frac{2 \sqrt{3}}{3}$$ and $$\frac{2 \sqrt{3}}{3}$$.
Sol.(iii) Consider equation $$2 x^2-6 x+3=0$$
Here, $$\quad a=2, b=-6, c=3$$
$$
\begin{aligned}
\quad \mathrm{D}=b^2-4 a c & =(-6)^2-4 \times 2 \times 3 \\
& =36-24=12>0
\end{aligned}
$$
As $$\mathrm{D}>0$$, therefore, roots are real and unequal.
Roots are $$x=\frac{-(-6) \pm \sqrt{12}}{4}=\frac{6 \pm 2 \sqrt{3}}{4}=\frac{3 \pm \sqrt{3}}{2}$$, i.e., $$\frac{3+\sqrt{3}}{2}$$ and $$\frac{3-\sqrt{3}}{2}$$.
