Question

Solve the simultaneos quadratic inequations. $x^{2}+5x+4>0$ and$-(x)^{2}-x-42>0$​

Answer 1

GIVEN:

★Two simultaneous quadratic inequations.

→$x^{2}+5x+4>0$

→$-(x)^{2}-x+42>0$

TO FIND:

★The range of values of x.

CONCEPT USED:

★We will solve the inequalities simultaneous and find out the critical points.

ANSWER:

Taking first equation,

=>$x^{2}+5x+4>0$

on factorising,

=>$x^{2}+x+4x+4>0$

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

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

. °. $x>-1, x>-4$

On drawing number line,

<--+ve------><----(-ve) ------><------+ve---->

<——————|————————|—————>

$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:-4\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: -1$

We would have to look for positive regions,

So,

$\large\green{\boxed{x\epsilon(-∞,-4) U(-1,∞)}}$.........(1)

________________________________________

Again,

=>$-(x)^{2}-x+42>0$

=>$-(x^{2}+x-42) >0$

=>$x^{2}+x-42) <0$

=>$x^{2}+7x-6x-42<0$

=>$x(x+7) -6(x+7) <0$

=>$(x+7) (x-6) <0$

On drawing number line,

<--+ve------><----(-ve) ------><------+ve---->

<——————|————————|—————>

$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:-7\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: 6$

Hence,

$\large\red{\boxed{x\epsilon(-7, 6)}}$ .............. (2)

________________________________________

From (1)&(2),we have;

$x\epsilon(-7,-4)U(-1,6)$

$\huge\orange{\boxed{x\epsilon(-7,-4)U(-1,6)}}$

Answer 2

Answer in attachment........,.....