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