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