Question

NEED THE ANSWERS FAST PLEASE

In Exercises 13 and 14, find a possible pair of integer values for a and c so that
the quadratic equation has the given number and type of solution(s). Then write
the equation.
13. ax² − 3x + c = 0; two real solutions
-
14. ax² + 10x + c = 0; two imaginary solutions
15. Determine the number and type of solutions to the equation 2x² - 8x = -15.
A. two real solutions
B. one real solution
C. two imaginary solutions
D. one imaginary solution
In Exercises 16 and 17, use the Quadratic Formula to write a quadratic equation
that has the given solutions.
16. x
10 ± √√-68
14
17. x =
-3±i√√7
8
In Exercises 18-21, solve the quadratic equation using the Quadratic Formula.
Then solve the equation using another method. Which method do you prefer?
Explain.
18. 7x² + 7 = 14x
20. x² + 2 = -x
19. x² + 20x = 8
21. 8x² - 48x + 64 = 0
22. The quadratic equation x² + x + c = 0 has two imaginary solutions. Show that
the constant c must be greater than 1.

Answer 1

Explanation:

the degree of a polynomial determines 1:1 the number of solutions.

a quadratic equation (degree 2) has 2 solutions.

the general solution is

x = (-b ± sqrt(b² - 4ac))/(2a)

in our case

a = -4

b = -4

c = -1

so,

x = (4 ± sqrt((-4)² - 4×-4×-1))/(2×-4)

when we look at the square root

16 - 16

we see that it is 0.

the square root of 0 is 0, and there is no difference between -0 and +0.

so, we get only one (real) solution : 4/-8 = -1/2

but : formally, there are still 2 solutions (as this is a quadratic equation). they are just identical.

so, I am not sure what your teacher wants to see in this case as answer.

my answer would be 2 real identical solutions. did this help?