Question

Solve the inequality :
$x^2+3x-4>0$

• A. $(-4,1)$
• B. $(-\infty ,-4)\cup (1,+\infty )$
• C. $(1,+\infty )$
• D. no solutions
• E. $(-\infty ,-4]\cup [1,+\infty )$

Answer 1

The correct option is B $(-\infty ,-4)\cup (1,+\infty )$

Finding the roots of the given equation as shown below:

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

$x^2+4x-x-4>0$

$x(x+4)-1(x+4)>0$

$(x+4)(x-1)>0$

Case-1:- Case-2:-

$x-1>0$ and $x+4>0$ $x-1<0$ and $x+4<0$

$x>1$ and $x>-4$ $x<1$ and $x<-4$

from above two we get from above two we get

$x>1$ $\dots \left(1\right)$ $x<-4$ $\dots \left(2\right)$

From $\left(1\right)$ and $\left(2\right)$

$x$ lies between $(-\infty ,-4)\cup (1,+\infty )$.

Hence option B is correct.