Multiple choice

Find the value of $\lambda$ for which the quadratic equation $3{ x }^{ 2 }+2({ \lambda }^{ 2 }+1)x+({ \lambda }^{ 2 }-3\lambda +2)=0$ has roots of opposite signs

  1. $1< \lambda < 2$
  2. $1< \lambda < 7$
  3. $1< \lambda < 8$

  4. None of these
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A Correct answer
Explanation

For a quadratic equation ax^2 + bx + c = 0 to have roots of opposite signs, the product of roots c/a must be less than 0. So (lambda^2 - 3*lambda + 2) / 3 < 0. (lambda - 1)(lambda - 2) < 0. This holds when 1 < lambda < 2.

AI explanation

For the roots of a quadratic equation to be opposite in sign, the product of the roots must be negative, which means the constant term divided by the leading coefficient must be less than zero. Here, the product of the roots is (lambda^2 - 3lambda + 2) divided by 3, so we set lambda^2 - 3lambda + 2 < 0. Factoring the quadratic gives (lambda - 1)(lambda - 2) < 0, which holds true when lambda lies strictly between 1 and 2. The value of lambda must satisfy 1 < lambda < 2.

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