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