Question

Find the roots of quadratic equation -

$2 {x}^{2} - ( \sqrt{26} - 1)x + 1$
Note -
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Answer 1

heya folk..!!

the given equation is 2x^2-(√26-1)x+1=0

comparing it with ax^2+bx+c=0.

we get a=2,b=-(√26-1) c=1

so, 'Discriminat ""D'"=(b^2-4*a*c)

=)D=b^2-4ac

=)(√26-1)^2-4*2*1

=)26+1-2√26*1-8

=)19-2√26>0
so the given equation has real roots

now, √D=√19-√2√√26

alpha,x=-b+√D

x=-2+√19-√2√√26/2*2
x=-2+-√19-√2√√26*/4

and again 2nd root will be.
bita =+b-√D/2a

bita /x=+2-(√19-2√√26)/4

x=2-√19+√2√√26/4

hope it help you.

@rajukumar☺