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प्रश्न
`x^2-2ax+(a^2-b^2)=0`
उत्तर
Given:
`x^2-2ax+(a^2-b^2)=0`
On comparing it with `Ax^2+Bx+C=0` we get
`A=1, B=-2a and C=(a^2-b^2)`
Discriminant D is given by:
`D=B^2-4AC`
=`(-2a)^2-4xx1xx(a^2-b^2)`
=`4a^2-4a^2+4b^2`
=`4b^2>0`
Hence, the roots of the equation are real.
Roots α and β are given by:
`α=(-b+sqrt(D))/(2a)=(-(-2a)+sqrt4b^2)/(2xx1)=(2a+2b)/2=(2(a+b))/2=(a+b)`
`β=(-b-sqrt(D))/(2a)=(-(-2a)-sqrt4b^2)/(2xx1)=(2a-2b)/2=(2(a-b))/2=(a-b)`
Hence, the roots of the equation are (a +b) and (a +b).
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