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rss GMAT Practice Question (One of Hundreds)
Which of the following equations has at least one root in common with x2 + 11x + 10 = 0?
Correct Answer: D
You only need to solve one equation
  1. First find the roots (or solutions) to the original equation.
    • Option 1: Factor the equation into (x + 10)(x + 1) = 0.
      This means that the solutions to the equation (or roots) are x = -10 and -1.
    • Option 2: Plug the equation into the quadratic formula with a=1, b=11, and c=10.
      When worked out, this also reveals x = -10 or -1.
  2. It is possible to proceed by evaluating the answer choices in the same manner as step one then check which option has a root in common with the original equation. However, a faster approach is to simply plug the two solutions found in step one into each equation.
  3. Plug x= -10 and -1 into A:
    (-10)2 - 11(-10) + 10 = 100+110+10 = 220.
    (-1)2 - 11(-1) + 10 = 1+11+10 = 22.
    Neither x = -10 nor x = -1 makes the equation equal to 0, so A can be ruled out.
  4. Plug x = -10 and -1 into B:
    (-10)2 + 10(-10) - 11 = 100-100-11 = -11.
    (-1)2 + 10(-1) - 11 = 1-10-11 = -20.
    Neither x = -10 nor x = -1 makes the equation equal to 0, so B can be ruled out.
  5. Plug x = -10 and -1 into C:
    (-10)2 + 10(-10) + 11 = 100-100+11 = 11
    (-1)2 + 10(-1) + 11 = 1-10+11 = 2
    Neither x = -10 nor x = -1 makes the equation equal to 0, so C can be ruled out.
  6. Plug x = -10 and -1 into D:
    2(-10)2 + (-10) - 1 = 2(100)-10-1 = 189
    2(-1)2 + (-1) - 1 = 2(1)-1-1 = 0
    x = -1 makes the equation equal to 0, so D is the answer, though it doesn't hurt to check the last answer to make sure.
  7. Plug x = -10 and -1 into E:
    2(-10)2 - (-10) - 1 = 2(100)+10-1 = 209
    2(-1)2 - (-1) - 1 = 2(1)+1-1 = 2
    Neither x = -10 nor x = -1 makes the equation equal to 0, so E can be ruled out.
  8. After reviewing all the choices, only D has a common root with the original equation.

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