Carcass wrote:
\(2x^2 + 6 > 40\)
Which values of x satisfy the inequality above?
Indicate all such values.
A. –8
B. –6
C. –4
D. –2
E. 2
F. 4
G. 6
H. 8
Given: \(2x^2 + 6 > 40\)
Subtract 6 from both sides: \(2x^2 > 34\)
Divide both sides by 2: \(x^2 > 17\)
This tells us that EITHER \(x > \sqrt{17}\) OR \(x < -\sqrt{17}\)
ASIDE: \(\sqrt{16}=4\) and \(\sqrt{25}=5\)
Since 17 is BETWEEN 16 and 25, we know that \(\sqrt{17}\) is BETWEEN 4 and 5.
So we can say, \(\sqrt{17}=\)
4.somethingWhen we applied this to her earlier solution, we can say that
EITHER x > 4.something OR x < -4.somethingWhen we examine the answer choices we can see that -8, -6, 6 and 8 satisfy those conditions.
Answer: A, B, G, H
Cheers,
Brent