The equations $x^{2} + x + a = 0$ and $x^{2} + ax + 1 = 0$ have a common real root
- For no value of $a$
- For exactly one value of $a$
- For exactly two values of $a$
- For exactly three values of $a$
Reveal answer
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B
Correct answer
Explanation
Subtracting the equations: (x^2 + x + a) - (x^2 + ax + 1) = 0 => x(1-a) - (1-a) = 0 => (x-1)(1-a) = 0. Either x=1 or a=1. If x=1, 1+1+a=0 => a=-2. If a=1, x^2+x+1=0 (no real roots). Thus, only a=-2 works.
AI explanation
Let the common root be x, so it satisfies both x^2 + x + a = 0 and x^2 + ax + 1 = 0. Subtracting the second equation from the first eliminates x^2 and gives x + a - ax - 1 = 0. Factoring out x from the terms gives (1 - a)x = 1 - a. Assuming a is not equal to 1, we divide both sides by (1 - a) to find the common root x = 1. Substituting x = 1 back into the first equation yields 1 + 1 + a = 0, so a = -2. Therefore, there is exactly one value of a, which is -2.