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प्रश्न
`x^2-(2b-1)x+(b^2-b-20)=0`
उत्तर
We write, -(2b-1)x=-(b-5)x-(b+4)x as
x^2xx(b^2-b-20)=(b^2-b-20)x^2=[-(b-5)x]xx[-(b+4)x]
∴ `x^2-(2b-1)x+(b^2-b-20)=0`
⇒ `x^2-(b-5)x-(b+4)x+(b+4)=0`
⇒`x[x-(b-5)]-(b+4)[x-(b-5)]=0`
⇒` [x-b-5]x- [b+4]=0`
⇒`x-(b-5)=0 or x-(b+4)=0`
⇒ `x=b-5 or x=b+4`
Hence, b - 5 and b + 4 are the roots of the given equation.
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