# Find the discriminant of each of the following equations:

Question:

Find the discriminant of each of the following equations:

(i) $2 x^{2}-7 x+6=0$

(ii) $3 x^{2}-2 x+8=0$

(iii) $2 x^{2}-5 \sqrt{2} x+4=0$

(iv) $\sqrt{3} x^{2}+2 \sqrt{2} x-2 \sqrt{3}=0$

(v) $(x-1)(2 x-1)=0$

(vi) $1-x=2 x^{2}$

Solution:

(i) $2 x^{2}-7 x+6=0$

Here,

$a=2$

$b=-7$

$c=6$

Discriminant $D$ is diven by :

$D=b^{2}-4 a c$

$=(-7)^{2}-4 \times 2 \times 6$

$=49-48$

$=1$

(ii) $3 x^{2}-2 x+8=0$

Here,

$a=3$,

$b=-2$,

$c=8$

Discriminant $D$ is given by :

$D=b^{2}-4 a c$

$=(-2)^{2}-4 \times 3 \times 8$

$=4-96$

$=-92$

(iii) $2 x^{2}-5 \sqrt{2 x}+4=0$

Here,

$a=2$

$b=-5 \sqrt{2}$

$c=4$

Discriminant $D$ is given by:

$D=b^{2}-4 a c$

$=(-5 \sqrt{2})^{2}-4 \times 2 \times 4$

$=(25 \times 2)-32$

$=50-32$

$=18$

(iv) $\sqrt{3} x^{2}+2 \sqrt{2} x-2 \sqrt{3}=0$

Here,

$a=\sqrt{3}$

$b=2 \sqrt{2}$

$c=-2 \sqrt{3}$

Discriminant $D$ is given by:

$D=b^{2}-4 a c$

$=(2 \sqrt{2})^{2}-4 \times \sqrt{3} \times(-2 \sqrt{3})$

$=(4 \times 2)+(8 \times 3)$

$=8+24$

$=32$

(v) $(x-1)(2 x-1)=0$

$\Rightarrow 2 x^{2}-3 x+1=0$

Comparing it with $a x^{2}+b x+c=0$, we get

$a=2, b=-3$ and $c=1$

$\therefore$ Discriminant, $D=b^{2}-4 a c=(-3)^{2}-4 \times 2 \times 1=9-8=1$

(vi) $1-x=2 x^{2}$

$\Rightarrow 2 x^{2}+x-1=0$

Here,

$a=2$,

$b=1$,

$c=-1$

Discriminant $D$ is given by:

$D=b^{2}-4 a c$

$=1^{2}-4 \times 2(-1)$

$=1+8$

$=9$