x²+x>1 in equality i cnat figure out how to factor it it does not make sense to me how you can have a single x and a negative 1
Hello,
$\displaystyle x^2+x > 1 \Longleftrightarrow x^2+x-1 > 0$
Now factor the LHS of this inequality:
$\displaystyle x^2+x-1 > 0 \Longleftrightarrow x^2+x+\frac{1}{4}-\frac{1}{4}-1>0 \Longleftrightarrow \left(x+\frac{1}{2} \right)^2-\frac{5}{4}>0$
$\displaystyle \left(x+\frac{1}{2} \right)^2-\frac{5}{4}>0 \Longleftrightarrow \left(x+\frac{1}{2} +\frac{1}{2} \cdot \sqrt{5} \right) \cdot \left(x+\frac{1}{2}-\frac{1}{2} \cdot \sqrt{5} \right)>0$
A product of 2 factors is positive (that means it is greater than zero) if both factors have the same sign. Therefore you get 2 sets of 2 inequalities:
$\displaystyle \left(x+\frac{1}{2} +\frac{1}{2} \cdot \sqrt{5} \right)>0\ \wedge \ \left(x+\frac{1}{2}-\frac{1}{2} \cdot \sqrt{5} \right)>0$ $\displaystyle \vee$
$\displaystyle \left(x+\frac{1}{2} +\frac{1}{2} \cdot \sqrt{5} \right)<0\ \wedge \ \left(x+\frac{1}{2}-\frac{1}{2} \cdot \sqrt{5} \right)<0$
Solve each inequality for x. Simplify. YOU'll get:
$\displaystyle x>-\frac{1}{2}+\frac{1}{2} \cdot \sqrt{5} \ \ \vee \ \ x< -\frac{1}{2} -\frac{1}{2} \cdot \sqrt{5}$
I've attached a diagram to show you how to solve such inequalities by graphing.
EB