Math - Mathematical Relationships and Basic Graphs

Solve for :

$(x+5)^{-3} = -1$

Explanation

Raise both sides of the equation to the inverse power of to cancel the exponent on the left hand side of the equation.

$\rightarrow ((x+5)^{-3})^{-\frac{1}{3}} = (-1)^{-\frac{1}{3}}$

$\rightarrow x+5 = -1$

Subtract from both sides:

$\rightarrow (x+5) - 5 = (-1)-5$

$\rightarrow x = -6$

Solve for :

$\left | 2x-5\right |=3x$

Explanation

To solve absolute value equations, we must understand that the absoute value function makes a value positive. So when we are solving these problems, we must consider two scenarios, one where the value is positive and one where the value is negative.

$\left | 2x-5\right |=3x$

and

This gives us:

and

$x=-5\ and\ 1$

However, this question has an outside of the absolute value expression, in this case . Thus, any negative value of will make the right side of the equation equal to a negative number, which cannot be true for an absolute value expression. Thus, is an extraneous solution, as $\left | 2x-5\right |$ cannot equal a negative number.

Our final solution is then

Which of the following is equivalent to $\frac{3+i}{3-4i}$ , where $i=\sqrt{-1}$ ?

$\frac{1}{5}+\frac{3}{5}i$

$\frac{1}{5}-\frac{3}{5}i$

$\frac{3+i}{3+4i}$

$\frac{3-i}{3-4i}$

Explanation

We will need to simplify the rational expression by removing imaginary terms from the denominator. Often in such problems, we want to multiply the numerator and denominator by the conjugate of the denominator, which will usually eliminate the imaginary term from the denominator.

In this problem, the denominator is . Remember that, in general, the conjugate of the complex number is equal to , where a and b are both nonzero constants. Thus, the conjugate of is equal to .

We need to multiply both the numerator and denominator of the fraction $\frac{3+i}{3-4i}$ by .

$\frac{3+i}{3-4i}=\frac{3+i}{3-4i}\cdot\frac{3+4i}{3+4i}=\frac{(3+i)(3+4i)}{(3-4i)(3+4i)}$

Next, we use the FOIL method to simplify both the numerator and denominator. The FOIL method of binomial multiplication requires us to mutiply together the first, outside, inner, and last terms of each binomial and then sum them.

$\frac{(3+i)(3+4i)}{(3-4i)(3+4i)}=\frac{3(3)+3(4i)+i(3)+i(4i)}{3(3)+3(4i)-4i(3)+(-4i)(4i)}$

Now, we can start simplifying.

$\frac{3(3)+3(4i)+i(3)+i(4i)}{3(3)+3(4i)-4i(3)+(-4i)(4i)}=\frac{9+12i+3i+4i^2}{9+12i-12i-16i^2}$

Use the fact that $i^2=(\sqrt{-1})^2=-1$ .

$\frac{9+12i+3i+4i^2}{9+12i-12i-16i^2}=\frac{9+12i+3i+4(-1)}{9+12i-12i-16(-1)}$

$=\frac{9+15i-4}{9+16}=\frac{5+15i}{25}=\frac{5}{25}+\frac{15i}{25}$

$=\frac{1}{5}+\frac{3}{5}i$

The answer is $\frac{1}{5}+\frac{3}{5}i$ .

Simplify the following complex number expression:

$(\sqrt{-6})(\sqrt{-4})(\sqrt{-3})$

$-6i\sqrt{2}$

$-6i\sqrt{3}$

$-4i\sqrt{3}$

$-4i\sqrt{2}$

$-6i\sqrt{5}$

Explanation

Begin by simplifying the radicals using complex numbers:

$(\sqrt{-6})(\sqrt{-4})(\sqrt{-3})$

$=(i\sqrt{6})(2i)(i\sqrt{3})$

Multiply the factors:

$=(2i^2\sqrt{6})(i\sqrt{3})$

$=(2i^3\sqrt{18})$

Simplify. Remember that is equivalent to .

$=-2i\sqrt{18}$

$=-6i\sqrt{2}$

Solve for :

$\sqrt{x^2+12x+36} =4$

$x = \left { -10, -2\right }$

Explanation

To solve for in the equation $\sqrt{x^2+12x+36} =4$

Square both sides of the equation

$\rightarrow (\sqrt{x^2+12x+36})^2 =(4)^2$

$\rightarrow x^2+12x+36 =16$

Set the equation equal to by subtracting the constant from both sides of the equation.

$\rightarrow (x^2+12x+36) - 16 =(16) -16$

$\rightarrow x^2+12x+20 =0$

Factor to find the zeros:

$\rightarrow (x+10)(x+2) = 0$

This gives the solutions

$x = \left { -10, -2\right }$ .

Verify that these work in the original equation by substituting them in for . This is especially important to do in equations involving radicals to ensure no imaginary numbers (square roots of negative numbers) are created.

Solve for :

$(x+5)^{-3} = -1$

Explanation

Raise both sides of the equation to the inverse power of to cancel the exponent on the left hand side of the equation.

$\rightarrow ((x+5)^{-3})^{-\frac{1}{3}} = (-1)^{-\frac{1}{3}}$

$\rightarrow x+5 = -1$

Subtract from both sides:

$\rightarrow (x+5) - 5 = (-1)-5$

$\rightarrow x = -6$

$y=\frac{3x^2-5}{2x^2+7}$

What are the horizontal asymptotes of this equation?

$y=\frac{3}{2}$

There are no horizontal asymptotes.

$y=\frac{2}{3}$

Explanation

Since the exponents of the variables in both the numerator and denominator are equal, the horizontal asymptote will be the coefficient of the numerator's variable divided by the coefficient of the denominator's variable.

For this problem, since we have $\frac{3x^2}{2x^2}$ , our asymptote will be $y=\frac{3}{2}$ .

$y=\frac{x^2-9}{x^2-16}$

What is the horizontal asymptote of this equation?

There is no horizontal asymptote.

Explanation

Look at the exponents of the variables. Both our numerator and denominator are . Therefore the horizontal asymptote is calculated by dividing the coefficient of the numerator by the coefficient of the denomenator.