Multiple choice

If one of the roots of the equation ${x}^{2}+ax+b=0$ and ${x}^{2}+bx+a=0$ is incident, then the numerical value of $(a+b)$ is

  1. $0$
  2. $-1$
  3. $-2$
  4. $5$
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B Correct answer
Explanation

If r is a common root, subtracting the two equations gives (a - b)(x - 1) = 0. For the intended non-identical equations, x = 1, and substitution gives 1 + a + b = 0. Therefore, a + b = -1.

AI explanation

Let alpha be the common root, so it satisfies both equations, meaning alpha^2 + a*alpha + b = 0 and alpha^2 + b*alpha + a = 0. Subtracting the second equation from the first gives (a - b)*alpha + (b - a) = 0, which factors to (a - b)(alpha - 1) = 0. Assuming a and b are distinct coefficients means alpha must be 1. Substituting alpha = 1 back into the first equation yields 1 + a + b = 0, which means a + b = -1.

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