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x(xa)b(a+b)=0x(x-a)-b(a+b)=0\newlineIn the given equation, aa and bb are constants. What are the solutions to the equation?\newlineChoose 11 answer:\newline(A) x=bx=-b and x=(ab)x=(a-b)\newline(B) x=bx=-b and x=(a+b)x=(a+b)\newline(C) x=ax=a and x=(ab)x=(a-b)\newline(D) x=ax=a and x=(a+b)x=(a+b)

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Q. x(xa)b(a+b)=0x(x-a)-b(a+b)=0\newlineIn the given equation, aa and bb are constants. What are the solutions to the equation?\newlineChoose 11 answer:\newline(A) x=bx=-b and x=(ab)x=(a-b)\newline(B) x=bx=-b and x=(a+b)x=(a+b)\newline(C) x=ax=a and x=(ab)x=(a-b)\newline(D) x=ax=a and x=(a+b)x=(a+b)
  1. Expand and Simplify Equation: First, we need to expand the equation to see if it can be factored easily.\newlinex(xa)b(a+b)=0x(x-a) - b(a+b) = 0\newlinex2axabb2=0x^2 - ax - ab - b^2 = 0
  2. Find Factors of Quadratic Equation: Now, we look for factors of the quadratic equation that could give us the solutions for xx. We need to find two numbers that multiply to give abb2-ab - b^2 and add up to give a-a.
  3. Analyze Factorization and Sum: We notice that the two numbers that satisfy these conditions are b-b and aba-b because:\newline(b)(ab)=ab+b2(-b) \cdot (a-b) = -ab + b^2\newline(b)+(ab)=b+ab=a2b(-b) + (a-b) = -b + a - b = a - 2b\newlineHowever, we need the sum to be a-a, not a2ba - 2b. This indicates that we might have made a mistake in our factorization or that the equation does not factor in the way we expected.

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