Problem 1 :
Solve the following quadratic equation using square root :
x2 = 36
Problem 2 :
Solve the following quadratic equation using square root :
x2 = -25
Problem 3 :
Find the sum of the following two complex numbers.
(3 - 4i) and (-8 + 6i)
Problem 4 :
Find the difference of the following two complex numbers.
(5 + 9i) and (3 - 4i)
Problem 5 :
Find the product in the form a + bi :
-3.5i(7 - 8i)
Problem 6 :
Find the product in the form a + bi :
(5 + 3i)(5 - 3i)
Problem 7 :
Write the following complex number in a + bi form :
5 / (2 - i)
Problem 8 :
Factor p2 + q2.
Problem 9 :
Factor 16p2 + 4.
Problem 10 :
Solve the following quadratic equation using factoring :
x2 + 16 = 0
Problem 1 :
Solve the following quadratic equation using square root :
x2 = 36
Solution :
In the given quadratic equation, x term is missing. So we can solve the equation using square root.
x2 = 36
Take square root on each side.
x = ±√36
x = ± 6
The solutions of the equation x2 = 36 are -6 and 6.
Problem 2 :
Solve the following quadratic equation using square root :
x2 = -25
Solution :
In the given quadratic equation, x term is missing. So we can solve the equation using square root.
x2 = -25
Take square root on each side.
x = ± √-25
x = ± √25√-1
x = ± 5i
The solutions of the equation x2 = -25 are not real numbers but are part of a number system called the complex numbers. The number √-1 is called the imaginary unit i. Replacing √-1 with i allows us to write the solutions to the equation x2 = -25 as 5i and -5i.
Problem 3 :
Find the sum of the following two complex numbers.
(3 - 4i) and (-8 + 6i)
Solution :
When adding two complex numbers in the form a + bi, combine the real parts and then combine the imaginary parts. The sum will include both a real and imaginary part and can be written in the form a + bi.
(3 - 4i) + (-8 + 6i) = [3 + (-8)] + [-4i + 6i]
= [3 - 8] + [2i]
= - 5 + 2i
Problem 4 :
Find the difference of the following two complex numbers.
(5 + 9i) and (3 - 4i)
Solution :
When subtracting two complex numbers in the form a + bi, combine the real parts and then combine the imaginary parts. The difference will include both a real and imaginary part and can be written in the form a + bi.
(5 + 9i) - (3 - 4i) = [5 - 3] + [9i - (-4i)]
= [2] + [9i + 4i]
= 2 + 13i
Problem 5 :
Find the product in the form a + bi :
-3.5i(7 - 8i)
Solution :
Use the distributive property.
-3.5i(7 - 8i) = -3.5i(7) - 3.5i(-8i)
Multiply.
= -24.5i + 28i2
Simplify using the definition i2.
= -24.5i + 28(-1)
= -24.5i - 28
Write in the form a + bi.
= -28 - 24.5i
Problem 6 :
Find the product in the form a + bi :
(5 + 3i)(5 - 3i)
Solution :
Use the distributive property.
(5 + 3i)(5 - 3i) = 5(5 - 3i) + 3i(5 - 3i)
Multiply.
= 25 - 15i + 15i - 9i2
= 25 - 9i2
Simplify using the definition i2.
= 25 - 9(-1)
= 25 + 9
= 34
Problem 7 :
Write the following complex number in a + bi form :
5 / (2 - i)
Solution :
When the denominator has an imaginary component, we can create an equivalent fraction with a real denominator by multiplying by its complex conjugate.
Use the complex conjugate of the denominator to multiply by 1.
5 / (2 - i) = [5 / (2 - i)] x 1
= [5 / (2 - i)] x [(2 + i) / (2 + i)]
= [5(2 + i)] / [(2 - i)(2 + i)]
Use the distributive property.
= (10 + 5i) / (4 + 2i - 2i - i2)
= (10 + 5i) / (4 - i2)
Simplify using the definition i2.
= (10 + 5i) / [4 - (-1)]
= (10 + 5i) / (4 + 1)
= (10 + 5i) / 5
= 10 / 5 + 5i /5
= 2 + i
Problem 8 :
Factor p2 + q2.
Solution :
Rewrite p2 + q2 as a difference of two squares :
p2 - (-q2)
We can think of (-q2) as
(-1)(q2)
Because -1 = i2, we have
(-q2) = (-1)(q2) = (i2)(q2) = (iq)2
So, we have
p2 + q2 = p2 - (-q2)
= p2 - (iq)2
= (p + iq)(p - iq)
Problem 9 :
Factor 16p2 + 4.
Solution :
Factor out the GCF :
16p2 + 4 = 4(4p2 + 1)
Rewrite as a difference of squares.
= 4(4p2 - i2)
Factor the difference of squares.
= 4(4p2 - i2)
= 4(4p + i)(4p - i)
The factors of 16p2 + 4 are 4, (4p + i) and (4p - i).
Problem 10 :
Solve the following quadratic equation using factoring :
x2 + 16 = 0
Solution :
Write the original equation :
x2 + 16 = 0
x2 + 42 = 0
Rewrite as a difference of squares.
x2 - (4i)2 = 0
(x + 4i)(x - 4i) = 0
By Zero Product Property,
x + 4i = 0, x - 4i = 0
x = -4i, x = 4i
The solutions are -4i and 4i.
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