Question

`2x^2-2sqrt2x+1=0` 

Answer 1

The given equation is `2x^2-2sqrt2x+1=0` 

Comparing it with `ax^2+bx+c=0` , we get 

`a=2, b=-2sqrt2 and  c=1` 

∴ Discriminant, `D=b^2-4ac=(-2sqrt2)^2-4xx2xx1=8-8=0` 

So, the given equation has real roots.
Now,  `sqrt(D)=0` 

∴ α= `(-b+sqrt(D))/(2a)=-(-2sqrt(2))/(2xx2)=(2sqrt(2))/4=sqrt(2)/2` 

β = `(-b-sqrt(D))/(2a)=(-(-2sqrt(2))-sqrt(0))/(2xx2)=(2sqrt(2))/4=sqrt(2)/2`  

Hence,`sqrt2/2` is the repeated root of the given equation.