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Bunuel
Which of the following equations has a root in common with x^2−7x+12=0 ?

A. x^2+1=0
B. x^2−x−12=0
C. x^2−8x−4=0
D. 3x^2−9=0
E. x^2+x-6=0

If we carefully look at the given equation we can arrange it in the following manner:
(x-3)(x-4) = 0
so the two roots are 3 and 4.

Now put 3 and 4 in given equations. The equation in which one of them gives value 0, that will be our answer.

(A) putting 3: we get 10; Putting 4 we get 17. Reject this option [by the way x^2+1 cannot have real roots]
(B) putting 3: we get 6; Putting 4 we get 0. This is the equation.

We are lucky . no need to check other options.

B is the answer.
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Bunuel
Which of the following equations has a root in common with x^2−7x+12=0 ?

A. x^2+1=0
B. x^2−x−12=0
C. x^2−8x−4=0
D. 3x^2−9=0
E. x^2+x-6=0

B is the correct answer

I believe the trick here is not solving the equation and it's roots but how to find the correct answer in the shortest time possible. I didn't substitute any numbers, I just looked at the answer choices and noticed that B can be solved by the same numbers (4&3) but with a different sign for (3).

I believe substitution for all the answer choices would be a waste of time, esp. that you have 2 numbers to substitute with (4&3)
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Bunuel
Which of the following equations has a root in common with x^2−7x+12=0 ?

A. x^2+1=0
B. x^2−x−12=0
C. x^2−8x−4=0
D. 3x^2−9=0
E. x^2+x-6=0

If two quadratic equations have a common root, then they have a common factor. Thus, we need to factor the left-hand side of the equation. Breaking down the given quadratic equation, we have:

(x - 4)(x - 3) = 0

We need to find an answer choice with at least one identical factor.

Looking at answer choice B we have:

x^2 − x − 12 = 0

(x - 4)(x + 3) = 0

Since answer choice B shares the factor (x - 4), both equations will have a common root, which is 4.

Answer: B
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