Question

If both roots of equation ax²+x+c-a=0are imaginary andc>-1 then
1)3a>2+4c 2)3a<2+4c 3)c<a 4)a>0​

Answer 1

Answer: a>0

Step-by-step explanation:

if roots are imaginary then D<0, where D= ${b^2-4ac$

using this,

$(1)^2-4a(c-a)$ < 0

$1-4ac+4a^2$ < 0

$1-c^2 + c^2-4ac+4a^2 <0\\1-c^2 +(c-2a)^2 <0\\(c-2a)^2<c^2-1$

now $(c-2a)^2 \geq 0$ (square can not be negative). if so, then we can conclude that $c^2-1$ must be positive (see above). for that to happen, c can either be greater than 1 or less than -1. but given that c>-1,  c must be greater than 1.

so, $c>1$ . now again consider

$1-4ac+4a^2 <0$     ( see step 2)

$1+4a^2 <4ac$    Now, $1 +4a^2 >0$ , so

$0 < 4ac\\0<ac$

It is only possible if

1.  a>0 and c>0   or

2. a<0 and c<0 . but option 2 is not possible because c>1.

so a must be greater than 0. that is $a>0$.

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