Question

1. what is the solution to $x^2 - 6 > 3x + 4$

fill in the blanks $x > \underline{5}$ or $x < \underline{-2}$

Explanation:

Step1: Rearrange to standard quadratic form

$x^2 - 3x - 6 - 4 > 0$
$x^2 - 3x - 10 > 0$

Step2: Factor the quadratic

$(x - 5)(x + 2) > 0$

Step3: Find critical points

$x - 5 = 0 \implies x = 5$
$x + 2 = 0 \implies x = -2$

Step4: Test intervals for inequality

Test $x < -2$: $(-)(-) = + > 0$ (valid)
Test $-2 < x < 5$: $(-)(+) = - < 0$ (invalid)
Test $x > 5$: $(+)(+) = + > 0$ (valid)

Answer:

$x>5$ or $x<-2$