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प्रश्न
`x^2-(sqrt2+1)x+sqrt2=0`
उत्तर
`x^2-(sqrt2+1)x+sqrt2=0`
⇒`x^2-(sqrt2+1)x=-sqrt2`
⇒`x^2-2xx x xx((sqrt2+1)/2)+((sqrt2+1)/2)^2=sqrt2+((sqrt2+1)/2)^2`
[Adding `((sqrt2+1)/2)]^2` on the sides]
⇒`[x-((sqrt2+1)/2)]^2=(-4sqrt2+2+1+2sqrt2)/4=(2-2sqrt2+1)/4=((sqrt2-1)/2)^2`
⇒`x-((sqrt2+1)/2)=+-((sqrt2-1)/2)` (Taking square root on both sides)
⇒`x-((sqrt2+1)/2)=((sqrt2-1)/2) or x-((sqrt2-1)/2)=-((sqrt2-1)/2)`
⇒`x=(sqrt2+1)/2+(sqrt2-1)/2 or x=(sqrt2+1)/2-(sqrt2-1)/2`
⇒`x=(2sqrt2)/2=sqrt2 or x=2/2=1`
Hence, `sqrt2 and 1` are the roots of the given equation.
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