Question

In the equation x^2-6x+c=0, find the values of c that will give two imaginary solutions.

a. c >= 5
b. c > 6
c. c < 6
d. c > 9

I figured out the equation with 6 (so I believe either b or c is the answer) being there, but is would it be > or <?

Rewrite x^2+9 as a product using complex numbers and the polynomial identity below.

a. (x+3)(x-3)
b. (x+3i)(x-3i)
c. (3x)(3i)
d. (x+9i)(x-9i)

I'm not sure if it's b or d, if I'm on the right path. Please help? Thanks

Answer 1

first one

when doing the quadratic equation, we come across something called the determinant

$\sqrt{b^2-4ac}$

when this is positive, we have 2 real roots
when this is 0, we have 1 real root
whenthis is negative, we have 2 imaginary rots
we want 2imaginary roots so
$\sqrt{b^2-4ac}$<0
we have
1x^2-6x+c=0
a=1
b=-6
c=c
$\sqrt{(-6)^2-4(1)(c)}$<0
$\sqrt{36-4c}$<0
square root both sies
36-4c<0
add 4c both sides
36<4c
divide both sides by 4
9<c
c>9

D is the answer (not b or c)


remember
√-1=i
a²-b²=(a-b)(a+b)

we want to do
x^2-(-9)
now we have to take the sqrt of -9
√-9=(√-1)(√9)=(i)(3)=3i

(x)^3-(3i)^2
(x-3i)(x+3i)
B is answer

Answer 2

The solution to the quadratic complex number problems are as follows;

1. Choice d: c > 9
2. Choice b: (x+3i)(x-3i)

In the quadratic root formular;

The expression b² - 4ac is the determinant of the kind of root the equation possesses;

In essence;

• If b² - 4ac > 0, the equation has 2 real roots

• If b² - 4ac = 0, the equation has 1 root

• If b² - 4ac < 0, the equation has 2 imaginary roots

Therefore we set;

• b² - 4ac < 0

where b = -6 and a =1.

• (-6)² - 4(1)×c < 0

• 36 - 4c < 0

• 36 < 4c

c > 9.

Therefore, the values of c that will give two imaginary solutions are: c > 9

Question 2:

To write x²+9 as a product using complex numbers and the polynomial identity;

we must first rewrite x²+9 as (x²- (-9))

By using the difference of two squares approach; we have;

• (x - 3i) (x + 3i)

Expansion of the expression above yields x² + 9.

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