Multiple choice

If $b < 0$, then the roots ${ x }{ 1 }$ and ${ x }{ 2 }$ of the equation $2{ x }^{ 2 }+6x+b=0$, satisfy the condition $\left( \dfrac { { x }{ 1 } }{ { x }{ 2 } } \right) +\left( \dfrac { { x }{ 2 } }{ { x }{ 1 } } \right) < K$, where $K$ is equal to

  1. $2$
  2. $-2$
  3. $0$
  4. $4$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given 2x^2 + 6x + b = 0. x1 + x2 = -3, x1*x2 = b/2. Expression = (x1/x2) + (x2/x1) = (x1^2 + x2^2) / (x1*x2) = ((x1+x2)^2 - 2x1x2) / (x1*x2) = (9 - b) / (b/2) = 18/b - 2. Since b < 0, 18/b is negative, so 18/b - 2 < -2.

AI explanation

Let x_1 and x_2 be the roots, so their sum is x_1 + x_2 = -3 and their product is x_1 * x_2 = b/2. The expression x_1/x_2 + x_2/x_1 simplifies to ((x_1 + x_2)^2 - 2*x_1*x_2) / (x_1*x_2). Substituting the known sum and product gives (-3)^2 / (b/2) - 2, which simplifies to 18/b - 2. Because b is strictly less than 0, the fraction 18/b is strictly negative, making the entire expression strictly less than -2.

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