Second-degree equations and inequalities via sign study

Text

Solve the following equations and inequalities, studying the sign of the associated parabola:

• (a) $(x+6)^2<1$
• (b) $\dfrac{1}{3}x^2+x\sqrt{7}+\dfrac{21}{4}\le 0$
• (c) $-x^2\le -3$
• (d) $-x^2+5x+9\le 0$

Solution

(a) $(x+6)^2<1 \iff |x+6|<1 \iff -7<x<-5$.

(b) Multiply by $12$: $4x^2+12\sqrt7\,x+63\le 0$. The discriminant is $\Delta=1008-1008=0$: the upward parabola touches the $x$-axis at a single point and is positive elsewhere. So it is $\le 0$ only at $x=-\dfrac{3\sqrt7}{2}$.

(c) $-x^2\le -3 \iff x^2\ge 3 \iff x\le -\sqrt3 \ \vee\ x\ge \sqrt3$.

(d) $-x^2+5x+9\le 0 \iff x^2-5x-9\ge 0$. Zeros $x=\dfrac{5\pm\sqrt{61}}{2}$; the upward parabola is $\ge 0$ outside them: $x\le \frac{5-\sqrt{61}}{2}\ \vee\ x\ge \frac{5+\sqrt{61}}{2}.$

$\boxed{\text{(a) } -7<x<-5;\ \ \text{(b) } x=-\tfrac{3\sqrt7}{2};\ \ \text{(c) } x\le-\sqrt3\vee x\ge\sqrt3;\ \ \text{(d) } x\le\tfrac{5-\sqrt{61}}{2}\vee x\ge\tfrac{5+\sqrt{61}}{2}}$