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प्रश्न
`36x^2-12ax+(a^2-b^2)=0`
उत्तर
The given equation is `36x^2-12ax(a^2-b^2)=0`
Comparing it with `Ax^2+Bx+C=0`
`A=36,B=-12a and C=a^2-b^2`
∴ Discriminant,
`D=B^2-4AC=(-12a)^2-4xx36xx(a^2-b^2)=144a^2-144a^2+144b^2=144b^2>0`
So, the given equation has real roots
Now, `sqrtD=sqrt144b^2=12b`
∴ `α =(-B+sqrtD)/(2A)=(-(-12a)+12b)/(2xx36)=(12(a+b))/72=(a+b)/0`
β=`(-B-sqrtD)/(2A)=(-(-12a)-12b)/(2xx36)=(12(a-b))/72=(a-b)/0`
Hence, `(a+b)/6` and `(a-b)/6` are the roots of the given equation.
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