Question

Find the value of discriminant.
(1) x² + 7x – 1 = 0
(2) 2y² – 5y + 10 = 0
(3) $\sqrt{2}{x}^2+4x+2\sqrt{2}=0$

Answer 1

(1) x² + 7x – 1 = 0
Comparing the given equation with $a{x}^2+bx+c=0$,
$a=1,b=7\mathrm{and}c=-1$
So, the discriminant $b^2-4ac={\left(7\right)}^2-4\times 1\times \left(-1\right)=49+4=53$

(2) 2y² – 5y + 10 = 0
Comparing the given equation with $a{x}^2+bx+c=0$,
$a=2,b=-5\mathrm{and}c=10$
So, the discriminant $b^2-4ac={\left(-5\right)}^2-4\times 2\times 10=25-80=-55$

(3) $\sqrt{2}{x}^2+4x+2\sqrt{2}=0$
Comparing the given equation with $a{x}^2+bx+c=0$,
$a=\sqrt{2},b=4\mathrm{and}c=2\sqrt{2}$
So, the discriminant $b^2-4ac={\left(4\right)}^2-4\times \sqrt{2}\times 2\sqrt{2}=16-16=0$