Inequalities

**pashah** · August 2nd 2006, 08:03 PM

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

x-3/x+4>7

x^2-5x-7<0

**earboth** · August 2nd 2006, 11:44 PM

Originally Posted by pashah

Solve the inequality. State the solution set using interval notation and graph.
x-3/x+4>7
x^2-5x-7<0

Hello, pashah,

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

Multiply by denominator. The next steps depends on the sign of the denominator!
$x-3>7x+28 \ \wedge\ x>-4$
$-31>6x \ \wedge\ x>-4$
$-\frac{31}{6}>x \ \wedge\ x>-4$ . that means $x\notin \mathbb{R}$
or
$x-3<7x+28 \ \wedge\ x<-4$
$-31<6x \ \wedge\ x<-4$
$-\frac{31}{6}<x \ \wedge\ x<-4$ . So the solution is:
$-\frac{31}{6}<x <-4$

2nd problem:
Solve this equation: $x^2-5x-7=0$ . You'll get:
$x=\frac{5}{2}-\frac{\sqrt{53}}{2}\ \vee\ x=\frac{5}{2}+\frac{\sqrt{53}}{2}$
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:
$\frac{5}{2}-\frac{\sqrt{53}}{2}<x<\frac{5}{2}+\frac{\sqrt{53}} {2}$

Greetings

EB

**earboth** · August 3rd 2006, 10:36 AM

Originally Posted by pashah

Solve the inequality. State the ... graph....

Hello, pashah,

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

Greetings

EB

**earboth** · August 3rd 2006, 10:59 AM

Originally Posted by pashah

Solve the inequality. State the ...graph.
x^2-5x-7<0

Hello, pashah,

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

Greetings

EB

Thread: Inequalities

LinkBack

Thread Tools

Display

Inequalities

Similar Math Help Forum Discussions

inequalities

inequalities

Inequalities

Inequalities

inequalities

Search Tags