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प्रश्न
` x^2+6x-(a^2+2a-8)=0`
उत्तर
We write,
`6x=(a+4)x-(a-2) x` as
`x^2xx[-(a^2a-8)]=-(a^2+2a-8)x^2=(a+4)x xx[-(a-2)x]`
∴`x^2+6x-(a^2+2a-8)=0`
⇒`x^2+(a+4)x-(a-2)x(a+4)(a-2)=0`
⇒`x[x+(a+4)]-(a-2)[x+(a+4)]=0`
⇒`[x+(a+4)][x-(a-2)]=0`
⇒`x+(a+4)=0 or x-(a-2)=0`
⇒`x=-(a+4) or x= a-2`
Hence, -(a+4) and (a-2) are the roots of the given equation.
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