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प्रश्न
`2x^2-2sqrt2x+1=0`
उत्तर
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.
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