Question

If $\frac{13x^2+30x+17}{3x^2+14x+8}>3,$ then find the value of x, if $xϵR$ and $x\ne \frac{-2}{3},-4.$

• A. $(-\infty ,-4)\cup (\frac{-2}{3},\frac{7}{2})\cup (\frac{9}{2},\infty )$
• B. $(-\infty ,-4)\cup (\frac{-2}{3},\frac{-1}{2})\cup (\frac{7}{2},\infty )$
• C. $(-4,\frac{-2}{3})\cup (\frac{-1}{2},\frac{7}{2})$
• D. $Noneofthese$

Answer 1

The correct option is B $(-\infty ,-4)\cup (\frac{-2}{3},\frac{-1}{2})\cup (\frac{7}{2},\infty )$

$\frac{13x^2+30x+17}{3x^2+14x+8}-3>0\frac{13x^2+30x+17-9x^2-42x-24}{3x^2+14x+8}>0\frac{4x^2-12x-7}{3x^2+14x+8}>0\frac{(2x+1)(2x-7)}{(3x+2)(x+4)}>0$
Multiplying both the numerator and the denominator by (3x + 2)(x+4), we get
(2x + 1)(2x -7)(3x + 2)(x + 4) > 0
(The sign does not change as $(3x+2{)}^2(x+4{)}^2$ is always positve).
Representing the above inequality on the number line we get,

The region where the inequality is satisfied is
$(-\infty ,-4)\cup (\frac{-2}{3},\frac{-1}{2})\cup (\frac{7}{2},\infty )$