Question

lf $x$ is real and $x^2-2x+3>0,12x^2+4x+3>0$, then $x$ lies in the interval

• A. $(2,4)$
• B. $(-\infty ,2)\cup (4,\infty )$
• C. $(-\infty ,\infty )$
• D. $[2,4]$

Answer 1

The correct option is C $(-\infty ,\infty )$

we have conditions,

Given,

$a{x}^2+bx+c>0$

if, $a>0,b^2-4ac<0$

Then, the interval of $x=(-\infty ,\infty )$

Given,

$x^2-2x+3>0$

$a=1>0$

$(-2{)}^2-4(1\left)\right(3)<0$.......(1)

and,

$12x^2+4x+3>0$

$a=12>0$

$(4{)}^2-4(12\left)\right(3)<0$......(2)

from (1) and (2), we have,

Then, the interval of $x=(-\infty ,\infty )$