Find the nature of the roots of the following quadratic equations:

Solution:

(i) The given equation is $2 x^{2}-8 x+5=0$.

This is of the form $a x^{2}+b x+c=0$, where $a=2, b=-8$ and $c=5$.

$\therefore$ Discriminant, $D=b^{2}-4 a c=(-8)^{2}-4 \times 2 \times 5=64-40=24>0$

Hence, the given equation has real and unequal roots.

(ii) The given equation is $3 x^{2}-2 \sqrt{6} x+2=0$.

This is of the form $a x^{2}+b x+c=0$, where $a=3, b=-2 \sqrt{6}$ and $c=2$.

$\therefore$ Discriminant, $D=b^{2}-4 a c=(-2 \sqrt{6})^{2}-4 \times 3 \times 2=24-24=0$

Hence, the given equation has real and equal roots.

(iii) The given equation is $5 x^{2}-4 x+1=0$.

This is of the form $a x^{2}+b x+c=0$, where $a=5, b=-4$ and $c=1$.

$\therefore$ Discriminant, $D=b^{2}-4 a c=(-4)^{2}-4 \times 5 \times 1=16-20=-4<0$

Hence, the given equation has no real roots.

(iv) The given equation is

$5 x(x-2)+6=0$

$\Rightarrow 5 x^{2}-10 x+6=0$

This is of the form $a x^{2}+b x+c=0$, where $a=5, b=-10$ and $c=6$

$\therefore$ Discriminant, $D=b^{2}-4 a c=(-10)^{2}-4 \times 5 \times 6=100-120=-20<0$

Hence, the given equation has no real roots.

(v) The given equation is $12 x^{2}-4 \sqrt{15} x+5=0$.

This is of the form $a x^{2}+b x+c=0$, where $a=12, b=-4 \sqrt{15}$ and $c=5$.

$\therefore$ Discriminant, $D=b^{2}-4 a c=(-4 \sqrt{15})^{2}-4 \times 12 \times 5=240-240=0$

Hence, the given equation has real and equal roots.

(vi) The given equation is $x^{2}-x+2=0$.

This is of the form $a x^{2}+b x+c=0$, where $a=1, b=-1$ and $c=2$.

$\therefore$ Discriminant, $D=b^{2}-4 a c=(-1)^{2}-4 \times 1 \times 2=1-8=-7<0$

Hence, the given equation has no real roots.