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प्रश्न
`4x^2-4bx-(a^2-b^2)=0`
उत्तर
The given equation is `4x^2-4bx-(a^2-b^2)=0`
Comparing it with `Ax^2+Bx+C=0` we get
A=4,B=4b and C=`-(a^2-b^2)`
`A=4,B=4b and C=-(a^2-b^2)`
∴ Discriminant,
`D=B^2-4AC=(4b)^2-4xx4xx[-(a^2-b^2)]=16b^2+16a^2-16b^2=16a^2>0`
So, the given equation has real roots
Now, `sqrtD=sqrt16a^2=4a`
∴` α=((-B+sqrt(D))/(2A))=(-4b+4a)/(2xx4)=(4(a-b))/8=(a-b)/2`
`β=((-B-sqrt(D))/(2A))=(-4b-4a)/(2xx4)=(4(a+b))/8=(a+b)/2`
Hence, `1/2(a-b)` and `-1/2(a+b)` are the roots of the given equation.
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