Multiple choice

If the equation ${ x }^{ 2 }+ix+a=0$, ${ x }^{ 2 }-2x+ia=0,a\neq 0$ have a common root, then

  1. $a$ is real
  2. $a=\dfrac{ 1 }{ 2 }+i$
  3. $a=\dfrac{ 1 }{ 2 }-i$

  4. The other root is also common
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let alpha be the common root. alpha^2 + i*alpha + a = 0 and alpha^2 - 2*alpha + i*a = 0. Subtracting: (i+2)alpha + a(1-i) = 0. alpha = -a(1-i)/(2+i). Substituting back into the first equation and solving for a yields a = 1/2 - i.

AI explanation

Let the common root of x^2 + ix + a = 0 and x^2 - 2x + ia = 0 be alpha. Multiplying the first equation by 2 and subtracting the second equation gives 2a - ia = 0. Since a is not 0, we divide by a to get 2 - i = 0, which means the common root alpha must be 2 - i. Substituting alpha = 2 - i into the first equation yields (2 - i)^2 + i(2 - i) + a = 0, which simplifies to 4 - 4i - 1 + 2i + 1 + a = 0, resulting in a = 1/2 - i.

Keep practicing — related questions