The value of $'a'$ for which the quadratic equation $2{x^2} - x\left( {{a^2} + 8a - 1} \right) + {a^2} - 4a = 0$ has roots opposite signs, lie in the interval

For a quadratic equation to have roots of opposite signs, the product of its roots must be less than zero, which means the constant term divided by the leading coefficient is negative. For the equation 2x^2 - x(a^2 + 8a - 1) + a^2 - 4a = 0, the product of the roots is (a^2 - 4a) / 2. We set the inequality (a^2 - 4a) / 2 < 0, which simplifies to a^2 - 4a < 0 or a(a - 4) < 0. Solving this inequality places 'a' in the interval 0 < a < 4.

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