Comment upon the nature of roots of the following quadratic equation.

1.	Comment upon the nature of roots of the following quadratic equation. (x-2a)(x-2b)=4ab\xa0
Answer» (x-2a)(x-2b)=4abx(x-2b)-2a(x-2b)=4abx2-2bx-2ax+4ab=4abx2-(2b+2a)x+4ab-4ab=0x2-(2a+2b)x+0=0So on comparing with quadratic eqaution Ax2+Bx+C=0We get A=1 B=-(2a+2b) C=0Discriminent D=B2-4AC =(-(2a+2b))2-4×1×0 =(2a+2b)2-0 =(2a+2b)2\xa0=(2(a+b))2=4(a+b)2Since square of any number cannot be negative, therefore D cannot be negative.So either D>0 or D=0Case1:- D=4(a+b)2>0 when a+b>0 , in this case roots are real and distinct.Case 2:- D=4(a+b)2=0 when\xa0a+b=0, in this case roots are real and repeated.

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