Find the interval(s) for b such that the following equation has at least one real solution
$x ^ { 2 } + b x + 6 = 0$ .

A) $( - \infty , - \sqrt { 24 } ] \cup [ \sqrt { 24 } , \infty )$
B) $( - \infty , - \sqrt { 24 } ] \cap [ \sqrt { 24 } , \infty )$
C) $( - \infty , \sqrt { 24 } ] \cup ( \sqrt { 24 } , \infty )$
D) $( - \infty , - \sqrt { 24 } ) \cup [ - \sqrt { 24 } , \infty )$
E) $( - \infty , \sqrt { 24 } ] \cap [ \sqrt { 24 } , \infty )$

Question

Find the interval(s) for b such that the following equation has at least one real solution 
$x ^ { 2 } + b x + 6 = 0$ .
 
A) $( - \infty , - \sqrt { 24 } ] \cup [ \sqrt { 24 } , \infty )$
B) $( - \infty , - \sqrt { 24 } ] \cap [ \sqrt { 24 } , \infty )$
C) $( - \infty , \sqrt { 24 } ] \cup ( \sqrt { 24 } , \infty )$
D) $( - \infty , - \sqrt { 24 } ) \cup [ - \sqrt { 24 } , \infty )$
E) $( - \infty , \sqrt { 24 } ] \cap [ \sqrt { 24 } , \infty )$

Quizplus · Accepted Answer

For a quadratic equation $ax^2 + bx + c = 0$ to have at least one real solution, the discriminant $b^2 - 4ac$ must be greater than or equal to 0. Here, $a = 1$, $b = b$, and $c = 6$. Thus, the discriminant is $b^2 - 4(1)(6) = b^2 - 24$. Setting this greater than or equal to 0 gives $b^2 \geq 24$, which means $b \leq -\sqrt{24}$ or $b \geq \sqrt{24}$, corresponding to the intervals $(-\infty, -\sqrt{24}]$ and $[\sqrt{24}, \infty)$.

Find the Interval(s) for B Such That the Following Equation x2+bx+6=0x ^ { 2 } + b x + 6 = 0x2+bx+6=0

Find the Interval(s) for B Such That the Following Equation $x ^ { 2 } + b x + 6 = 0$