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प्रश्न
`x^2-4ax-b^2+4a^2=0`
उत्तर
The given equation is `x^2-4ax-b^2+4a^2=0`
Comparing it with `Ax^2+Bx^+C=0`
`A=1,B=-4a and C=-b^2+4a`
∴ Discriminant, `D=B^2-4AC=(-4a)^2-4xx1xx(-b^2+4a^2)=16a^2+4b^2-16a^2=4b^>0`
So, the given equation has real roots
Now, `sqrtD=sqrt4b^2=2b`
∴ `α=(-B+sqrtD)/(2a)=(-(-4a)+2b)/(2xx1)=(4a+2b)/2=2a+b`
`β=-(-B-sqrtD)/(2a)=(-(-4a)-2b)/(2xx1)=(4a-2b)/2=2a-b`
Hence,` (2a+b) and (2a-b)` are the roots of the given equation.
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