Question

# If both the roots of the equation $$a{ x }^{ 2 }+2xa+1+{ a }^{ 2 }-16=0$$ are opposite in sign, then the range of $$a$$ is

A
(,4)(4,)
B
(4,4)
C
(,4)(0,4)
D
(0,4)

Solution

## The correct option is C $$\left( -\infty ,-4 \right) \cup \left( 0,4 \right)$$$$ax^{2}+2ax+1+a^{2}-16=0$$ [Ref. image 1]$$(a)f(0) < 0\Rightarrow (a)(1+a^{2}-16) < 0$$                           $$(a) (a+4)(a-4) < 0$$$$\therefore a\epsilon (-\infty ,-4) \cup (0,4)$$ [Ref. image 2]$$\therefore$$ option CMathematics

Suggest Corrections

0

Similar questions
View More

People also searched for
View More