156k views
3 votes
Consider this quadratic equation.

x²+2x+7=21
The number of positive solutions to this equation is
The approximate value of the greatest solution to the equation, rounded to the nearest hundredth, is

User Amuniz
by
6.7k points

1 Answer

4 votes

Answer: We can start by simplifying the equation:

x² + 2x + 7 = 21

x² + 2x - 14 = 0

Next, we can use the quadratic formula to solve for x:

x = (-b ± sqrt(b² - 4ac)) / 2a

where a = 1, b = 2, and c = -14

x = (-2 ± sqrt(2² - 4(1)(-14))) / 2(1)

x = (-2 ± sqrt(68)) / 2

x = (-2 ± 2sqrt(17)) / 2

x = -1 ± sqrt(17)

We have two solutions: x = -1 + sqrt(17) and x = -1 - sqrt(17). Since both solutions are real, there are no positive solutions to the equation.

To find the greatest solution to the equation, we can use the fact that the solutions are -1 + sqrt(17) and -1 - sqrt(17). The greatest solution is the one that is furthest to the right on the number line.

Since both solutions are negative, the greatest solution is the one with the smaller absolute value. So, the greatest solution is -1 - sqrt(17).

Rounding this to the nearest hundredth gives:

-1 - sqrt(17) ≈ -4.12

Therefore, the approximate value of the greatest solution to the equation, rounded to the nearest hundredth, is -4.12.

Explanation:

User PeSHIr
by
7.3k points