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प्रश्न
Solve` x^2-4ax+4a^2-b^2=0`
उत्तर
`x^2-4ax+4a^2-b^2=0`
⇒`x^2-4ax+(2a+b) (2a-b)=0`
⇒`x^2-[(2a+b)+(2a-b)]x+(2a+b) (2a-b)=0`
⇒`x^2-(2a+b)x-(2a-b)x+(2a+b)(2a-b)=0`
⇒`x[x-(2a+b)]-(2a-b)[x-(2a+b)=0]`
⇒`[x-(2a+b)][x-(2a-b)]=0`
⇒`x-(2a+b)=0 or x-(2a-b)=0`
⇒`x=(2a+b) or x=(2a-b)`
Hence, `(2a+b) and (2a-b)` are the roots of the given equation.
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