Solve the inequality. State the solution set using interval notation and graph.

x-3/x+4>7

x^2-5x-7<0

Printable View

- Aug 2nd 2006, 08:03 PMpashahInequalities
Solve the inequality. State the solution set using interval notation and graph.

x-3/x+4>7

x^2-5x-7<0 - Aug 2nd 2006, 11:44 PMearbothQuote:

Originally Posted by**pashah**

I assume that your first problem reads:$\displaystyle \frac{x-3}{x+4}>7$

Multiply by denominator. The next steps depends on the sign of the denominator!

$\displaystyle x-3>7x+28 \ \wedge\ x>-4$

$\displaystyle -31>6x \ \wedge\ x>-4$

$\displaystyle -\frac{31}{6}>x \ \wedge\ x>-4$. that means $\displaystyle x\notin \mathbb{R}$

or

$\displaystyle x-3<7x+28 \ \wedge\ x<-4$

$\displaystyle -31<6x \ \wedge\ x<-4$

$\displaystyle -\frac{31}{6}<x \ \wedge\ x<-4$. So the solution is:

$\displaystyle -\frac{31}{6}<x <-4$

2nd problem:

Solve this**equation:**$\displaystyle x^2-5x-7=0$. You'll get:

$\displaystyle x=\frac{5}{2}-\frac{\sqrt{53}}{2}\ \vee\ x=\frac{5}{2}+\frac{\sqrt{53}}{2}$

Because this equation belongs to a quadratic funtion which open upward, the part of the function between the zeros is below the x-axis, that means it is smaller than zero. Therefore your solution is:

$\displaystyle \frac{5}{2}-\frac{\sqrt{53}}{2}<x<\frac{5}{2}+\frac{\sqrt{53}} {2}$

Greetings

EB - Aug 3rd 2006, 10:36 AMearbothQuote:

Originally Posted by**pashah**

I've attached the graph of the 1rst problem.

Greetings

EB - Aug 3rd 2006, 10:59 AMearbothQuote:

Originally Posted by**pashah**

I've attached the graph of your 2nd problem.

Greetings

EB