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प्रश्न
`x^2+6x-(a^2+2a-8)=0`
उत्तर
The given equation is `x^2+6x-(a^2+2a-8)=0`
Comparing it with `Ax^2+Bx+C=0`
`A=1,B=6 and C=-(a^2+2a-8)`
∴ Discriminant, D=
`B^2-4AC=6^2-4xx1xx[-(a^2+2a-8)]=36+4a^2+8a-32=4a^2+8a-32=4a^2+8a+4`
`=4(a^2+2a+1)=4(a+1)^2>0`
So, the given equation has real roots
Now, `sqrtD=sqrt(4(a+1)^2)=2(a+1)`
∴`α=(-B+sqrt(D))/(2A)=(-6+2(a+1))/(2xx1)=(2a-4)/2=a-2`
`β=(-B-sqrt(D))/(2A)=(-6-2(a+1))/(2xx1)=(2a-8)/2=a-4=-(a+4)`
Hence, (a-2) and -(a+4) are the roots of the given equation.
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