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प्रश्न
`x^2+5x-(a^+a-6)=0`
उत्तर
The given equation is `x^2+5x-(a^2+a-6)=0`
Comparing it with `Ax^2+Bx+C=0`
`A=1,B=5 and C=-(a^2+a-6)`
∴ Discriminant, D =
`B^2-4AC=5^2-4xx1xx[-(a^2+a-6)]=25+4a^2+4a-24=4a^2+4a^2+4a+1`
=`(2a+1)^2>0`
So, the given equation has real roots
Now, `sqrtD=sqrt((2a+1))^2=2a+1`
∴`α=(-B+sqrtD)/(2A)=(-5+2a+1)/(2xx1)=(2a-4)/2=a-2`
`β=(-B-sqrtD)/(2A)=(-5-2a+1)/(2xx1)=(-2a-6)/2=a-2=-a-3=-(a+3)`
Hence, (a-2) and -(a+3) are the roots of the given equation.
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