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^{2}+ 11x + 10 = 0?

**D**

- 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.

- 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.
- 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. - 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. - 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. - 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. - 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. - After reviewing all the choices, only D has a common root with the original equation.

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