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

The discriminant of the quadratic equation is $b^2 - 4(1)(6) = b^2 - 24$. For the equation to have at least one real solution, the discriminant must be non-negative. Therefore, $b^2 - 24 \ge 0$, which is equivalent to $(b - \sqrt{24})(b + \sqrt{24}) \ge 0$. This inequality is satisfied when $b \le -\sqrt{24}$ or $b \ge \sqrt{24}$.

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

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