Solving a square inequality

I am addressing the adhering to inequality, please consider it and also inform me whether am I deal with or otherwise. This is an instance in Howard Anton's publication and also I addressed it on my very own as offered listed below, yet guide has actually addressed it in different ways! I intend to validate that my remedy is additionally legitimate.

0
2019-05-09 11:43:24
Source Share
Answers: 3

For abdominal muscle to be favorable either

  • an and also b are both favorable
  • an and also b are both adverse

Here, a = x - 5 and also b = x+2

They are both favorable if x > 5. They are both adverse if x < - 2. Either of these will certainly address the trouble

0
2019-05-10 10:14:57
Source

If you chart the function $y=x^2-3x-10$, you can see that the remedy is $x<-2$ or $x>5$.

0
2019-05-10 10:04:35
Source

Casebash's solution is great.

Below is a 2nd solution. You can use the adhering to

Theorem : If the origins $x_{1},x_{2}$ of $f(x)=ax^{2}+bx+c$ are actual and also $x_{1}\neq x_{2}$ (with $x_{1} < x_{2}$), after that, the signal of $f(x)$ is :

  • contrary to the signal of $a$ for $x\in \left[ x_{1},x_{2}\right] $,
  • the very same of $a$ for $x \ in \ left ] - \ infty, x _ 1 \ appropriate [ \ vee x \ in \ left ] x _ 2 , - \ infty \ right [ $.

Given that in your instance $a=1>0$, $x_{1}=-2<5=x_{2}$, you have $x^{2}-3x-10>0$ for $x\in \left] -\infty ,-2\right[ \vee x\in \left] 5,\infty \right[ $.

Addendum : A feasible evidence of this theory is to make use of the description of Casebash, thinking about that $ax^{2}+bx+c=a(x-x_1)(x-x_2)$

0
2019-05-10 10:03:08
Source