Question

# 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

A
1<a<5
B
0<a<4
C
1<a<2
D
2<a<6

Solution

## The correct option is B $$0 < a < 4$$R.E.F image Given equation is : $$2x^{2}-x(a^{2}+8a-1)+a^{2}-4a = 0$$Comparing it with the equation : $$ax^{2}+bx+c = 0$$we have $$a = 2$$$$b = -(a^{2}+8a-1)$$$$c = a^{2}-4a$$For the roots to have opposite signs.product of roots $$< 0$$$$\Rightarrow \dfrac{c}{a}< 0$$$$\Rightarrow \dfrac{a^{2}-4a}{2}< 0$$$$\Rightarrow a^{2}-4a< 0$$$$\Rightarrow a(a-4)< 0$$$$\Rightarrow (a-0)(a-4)< 0$$$$\therefore aE(0,4)$$ Mathematics

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