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3 years ago

2+3i divided by 3-3i ( a+bi form)

1 answer:

6 0

$(2+3i):(3-3i)=\frac{2+3i}{3-3i}\cdot\frac{3+3i}{3+3i}=\frac{(2+3i)(3+3i)}{(3-3i)(3+3i)}=\frac{6+6i+9i-9}{3^2-(3i)^2}=\frac{-3+15i}{9+9}\\\\=\frac{-3+15i}{18}=-\frac{3}{18}+\frac{15i}{18}=-\frac{1}{6}+\frac{5}{6}i$

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4 years ago

A positive real number is 4 less than another. If the sum of the squares of the two numbers is 72, then find the numbers.

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Answer:

Our two numbers are:

$2+4\sqrt{2} \text{ and } 4\sqrt{2}-2$

Or, approximately 7.66 and 3.66.

Step-by-step explanation:

Let the two numbers be a and b.

One positive real number is four less than another. So, we can write that:

$b=a-4$

The sum of the squares of the two numbers is 72. Therefore:

$a^2+b^2=72$

Substitute:

$a^2+(a-4)^2=72$

Solve for a. Expand:

$a^2+(a^2-8a+16)=72$

Simplify:

$2a^2-8a+16=72$

Divide both sides by two:

$a^2-4a+8=36$

Subtract 36 from both sides:

$a^2-4a-28=0$

The equation isn't factorable. So, we can use the quadratic formula:

$\displaystyle x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

In this case, a = 1, b = -4, and c = -28. Substitute:

$\displaystyle x=\frac{-(-4)\pm\sqrt{(-4)^2-4(1)(-28)}}{2(1)}$

Evaluate:

$\displaystyle x=\frac{4\pm\sqrt{128}}{2}=\frac{4\pm8\sqrt{2}}{2}=2\pm4\sqrt{2}$

So, our two solutions are:

$\displaystyle x_1=2+4\sqrt{2}\approx 7.66\text{ or } x_2=2-4\sqrt{2}\approx-3.66$

Since the two numbers are positive, we can ignore the second solution.

So, our first number is:

$a=2+4\sqrt{2}$

And since the second number is four less, our second number is:

$b=(2+4\sqrt{2})-4=4\sqrt{2}-2\approx 3.66$

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