Show all work to multiply quantity 2 plus the square root of negative 25 end quantity times quantity 4 minus the square root of negative 100 end quantity
The Correct Answer and Explanation is:
We are asked to multiply:(2+−25)(4−−100)(2 + \sqrt{-25})(4 – \sqrt{-100})(2+−25)(4−−100)
Step 1: Simplify the square roots of negative numbers
We use the imaginary unit i, where i=−1i = \sqrt{-1}i=−1.
- −25=25⋅−1=5i\sqrt{-25} = \sqrt{25} \cdot \sqrt{-1} = 5i−25=25⋅−1=5i
- −100=100⋅−1=10i\sqrt{-100} = \sqrt{100} \cdot \sqrt{-1} = 10i−100=100⋅−1=10i
So the expression becomes:(2+5i)(4−10i)(2 + 5i)(4 – 10i)(2+5i)(4−10i)
Step 2: Use the distributive property (FOIL)
We apply the FOIL method:
First:
2⋅4=82 \cdot 4 = 82⋅4=8
Outer:
2⋅(−10i)=−20i2 \cdot (-10i) = -20i2⋅(−10i)=−20i
Inner:
5i⋅4=20i5i \cdot 4 = 20i5i⋅4=20i
Last:
5i⋅(−10i)=−50i25i \cdot (-10i) = -50i^25i⋅(−10i)=−50i2
Step 3: Combine the results
Now add them:8−20i+20i−50i28 – 20i + 20i – 50i^28−20i+20i−50i2
Note that the imaginary terms cancel:−20i+20i=0-20i + 20i = 0−20i+20i=0
We are left with:8−50i28 – 50i^28−50i2
Step 4: Simplify i2i^2i2
We know:i2=−1i^2 = -1i2=−1
So:−50i2=−50(−1)=50-50i^2 = -50(-1) = 50−50i2=−50(−1)=50
Now add:8+50=588 + 50 = 588+50=58
Final Answer:
58\boxed{58}58
Explanation
In this problem, we are multiplying two complex-number-like expressions that involve square roots of negative numbers. Square roots of negative numbers do not exist in the real number system, but they are defined in the complex number system using the imaginary unit i, where i=−1i = \sqrt{-1}i=−1.
The first step is to simplify the square roots. For example, −25\sqrt{-25}−25 becomes 5i5i5i because 25=5\sqrt{25} = 525=5 and −1=i\sqrt{-1} = i−1=i. Similarly, −100=10i\sqrt{-100} = 10i−100=10i. Replacing these in the original expression transforms it into (2+5i)(4−10i)(2 + 5i)(4 – 10i)(2+5i)(4−10i).
To multiply two binomials, we use the distributive property, commonly known as the FOIL method. This method stands for First, Outer, Inner, and Last terms. Multiplying each of these term pairs gives us the components that make up the expanded expression.
Adding the results, we notice that the imaginary parts cancel out. This leaves us with two real number components: 8 and −50i2-50i^2−50i2. Remembering that i2=−1i^2 = -1i2=−1, we substitute and find that −50i2-50i^2−50i2 becomes 50. Adding 8 and 50 gives the final result of 58.
This shows how imaginary numbers can interact algebraically to produce real results. The key is understanding how to handle iii and applying basic algebra rules carefully.
