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If aa and bb are non-zero integers such that a2+b24a2b=0a^2+b^2-4a-2b=0 and a2b2=0a^2-b^2=0, which of the following could be the value of (ab)(a-b)?\newline(A) 3-3\newline(B) 22\newline(C) 33\newline(D) 66

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Q. If aa and bb are non-zero integers such that a2+b24a2b=0a^2+b^2-4a-2b=0 and a2b2=0a^2-b^2=0, which of the following could be the value of (ab)(a-b)?\newline(A) 3-3\newline(B) 22\newline(C) 33\newline(D) 66
  1. Identify Equations and Relationship: Identify the given equations and the relationship between aa and bb. We have two equations: 11. a2+b24a2b=0a^2 + b^2 − 4a − 2b = 0 22. a2b2=0a^2 − b^2 = 0 The second equation can be factored using the difference of squares. a2b2=(a+b)(ab)=0a^2 − b^2 = (a + b)(a − b) = 0 Since aa and bb are non-zero integers, a+ba + b cannot be zero, so aba − b must be zero.
  2. Determine Value of a: Determine the value of a in terms of b using the second equation.\newlineFrom a2b2=0a^2 - b^2 = 0, we have:\newlineab=0a - b = 0\newlinea=ba = b
  3. Substitute Value of aa: Substitute the value of aa into the first equation.\newlineSince a=ba = b, we can replace aa with bb in the first equation:\newlineb2+b24b2b=0b^2 + b^2 − 4b − 2b = 0\newlineCombine like terms:\newline2b26b=02b^2 − 6b = 0
  4. Factor Out bb: Factor out bb from the equation.b(2b6)=0b(2b − 6) = 0Since bb is a non-zero integer, we can't have b=0b = 0, so we must have:2b6=02b − 6 = 0
  5. Solve for b: Solve for b.\newlineAdd 66 to both sides of the equation:\newline2b=62b = 6\newlineDivide both sides by 22:\newlineb=3b = 3
  6. Find Value of a: Since a=ba = b, find the value of aa.\newlinea=b=3a = b = 3
  7. Calculate (ab)(a - b): Calculate the value of (ab)(a − b).\newlineSince a=b=3a = b = 3, we have:\newline(ab)=33=0(a − b) = 3 − 3 = 0\newlineHowever, this contradicts the options given, as none of them are zero. We need to recheck our steps for any errors.

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