Question

Please check me answer!
Which is true about the solutions to x2 + 2x−8=0?

a. No real solutions
b. Two identical rational solutions
c. Two different rational solutions
d. Two irrational solutions

Is the answer C correct?

Answer 1

To determine the types and number of roots of a
quadratic equation, examine the discriminant:
• If b² - 4ac > 0 then there are two real (unique) roots
• If b² - 4ac = 0 then there is one (double) real root
• If b² - 4ac < 0 then there are two imaginary roots

In this case: x^2 + 2x - 8 = 0
• (2)² - 4(1)(-8) = 36 > 0 thus there are two real (unique) roots
And since √36 = 6, both roots are rational.

Answer 2

x^2 + 2x−8=0
\[x = \frac{-2 ± \sqrt{2^2 - 4(1)(-8)}}{2(1)} = \frac{-2 ± \sqrt{36}}{2}=-1 ± 6\]
x = -1 + 6 = 5 or x = -1 - 6 = -7

Answer 3

Therefore the answer is c. Two different rational solutions

Answer 4

Thank you guys!