Let us consider a=b=2; and c=9 in the given equation (x-a)(x-b)=c So, (x-2)^2=9=>(x-2)=3 or (x-2)=-3 =>x=5 or x=-1; same as p=5 and q=-1 This comes down to 5(value of p) and -1(value of q) are the roots of the equation (x-a)(x-b)=c where a=2, b=2 and c=9; Now,
(x-a)(x-b)+c=0 gives (x-5)(x+1)+9=0 =>x^2-4x-5+9=0 =>x^2-4x+4=0 =>(x-2)^2=0 =>x=2 or x=-2 The roots of the given equation are 2 and -2 The roots contradict with all given options in terms of a, b, or c
May 27th, 2015
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