Question

Find the discriminant of each of the following equations :
(i) $2x^2-7x+6=0$ (ii) $3x^2-2x+8=0$
(iii) $2x^2-5\sqrt{2}x+4=0$ (iv) $\sqrt{3}{x}^2+2\sqrt{2}x-2\sqrt{3}=0$
(v) $(x-1)(2x-1)=0$ (vi) $1-x=2x^2$

Answer 1

(i) $2x^2-7x+6=0$
The given quadratic equation is in the form of
$a{x}^2+bx+c=0$
Where, $a=2,b=-7,c=6$
Thus, discriminant $D=b^2-4ac$
$=7^2-4\times 2\times 6$
$=49-48$
$\therefore D=1$
(ii) $3x^2-2x+8=0$
Comparing with $a{x}^2+bx+c=0$, we get
$a=3,b=-2,c=8$
Thus, the discriminant $D=b^2-4ac$
$=(-2{)}^2-(4\times 3\times 8)$
$=4-96$
$\therefore D=-92$

(iii) $2x^2-5\sqrt{2}x+4=0$
Where $a=2,b=-5\sqrt{2},c=4$
$D=b^2-4ac$
$=(-5\sqrt{2}{)}^2-4\times 2\times 4$
$=50-32$
$\therefore D=18$
iv) $\sqrt{3}{x}^2+2\sqrt{2}x-2\sqrt{3}=0$
$a=\sqrt{3},b=2\sqrt{2},c=-2\sqrt{3}$
$D=b^2-4ac$
$=(2\sqrt{2}{)}^2-4\times \surd 3\times -2\surd 3$
$=8+24$
$\therefore D=32$

(v)$(x-1)(2x-1)=0$
$2x^2-x-2x+1=0$
$2x^2-3x+1=0$
Where $a=2,b=-3,c=1$
D=$b^2-4ac$
$=(-3{)}^2-4\times 2\times 1$
$=9-8$
$\therefore D=1$

(vi) $1-x=2x^2$
Given equation is written as
$2x^2+x-1=0$
Where $a=2,b=1,c=-1$
$D=b^2-4ac$
$=1^2-(4\times 2\times -1)$
$=1+8$
$\therefore D=9$