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प्रश्न
`2x^2-x+1/8=0`
उत्तर
We Write, , `-x=-x/2-x/2` as `2x^2xx1/8=x^2/4=(-x/2)xx(-x/2)`
`∴2x^2-x+1/8=0`
⇒ `2x^2-x/2-x/2+1/8=0`
⇒`2x(x-1/4)-1/2(x-1/4)=0`
⇒`(x-1/4)(2x-1/2)=0`
⇒`x-1/4=0 or 2x-1/2=0`
`⇒x=1/4 or x=1/4`
Hence, `1/4` is the repeated root of the given equation.
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