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Question: The quadratic equations \(2{x^2} - ({a^3} + 8a - 1)x + {a^2} - 4a = 0\) possess roots of opposite si...

The quadratic equations 2x2−(a3+8a−1)x+a2−4a=02{x^2} - ({a^3} + 8a - 1)x + {a^2} - 4a = 0 possess roots of opposite sign then
A. a⩽0a \leqslant 0
B. 0<a<40 < a < 4
C. 4⩽a<84 \leqslant a < 8
D. a⩾8a \geqslant 8

Explanation

Solution

Hint: Here we will use the concept of product of roots and find the range of a in the quadratic equation.

Complete step-by-step answer:

Now we have been given a quadratic equation 2x2−(a3+8a−1)x+a2−4a=02{x^2} - ({a^3} + 8a - 1)x + {a^2} - 4a = 0
It’s been given that roots are of the opposite sign that one is negative and one is positive so their product should be negative obviously.
Formula for the product of roots is given as ca\dfrac{c}{a}and it should be negative.
Hence

\dfrac{c}{a} < 0 \\\ {\text{or }}\dfrac{{{a^2} - 4a}}{2} < 0 \Rightarrow {a^2} - 4a < 0 \\\ \end{gathered} $$ We can write this down as $a(a - 4) < 0$. This implies that $a \in (0,4)$. Hence the correct option is (b). Note: Whenever we come across such questions we simply need to recall the concept of sum and the product of roots, this concept helps reach the right answer.