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

1057567_1089130_ans_2599d84e7a984f939d5953335438b73b.png

Mathematics

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