Question

Please answer and show work/explain.

Answer 1

Answer:

5. x = { -13/2, 13/2 }

6. x = { (-5/4)±(i√(23)/4) }

7. x = { 1 - √7, 1 + √7 }

8. z = { -9i ,9i }

Step-by-step explanation:

Most quadratic equations can be graphed with the help of their x-intercepts. These x-intercepts, also known as zeroes or solutions, can be found by solving the quadratic equation. In a quadratic equation, when x is isolated, what x equals to is considered the solutions of the equation.

There are four main methods of solving quadratic equations: factoring, using the quadratic formula, using the square root method, and completing the square. I will try to solve each of these eqautions with one of the four methods.

5. 4x^2 - 169 = 0 (The Square Root Method)

Quadratic equations have a standard form of ax^2 + bx + c = 0, where a, b, and c are constants. The square root method can be used if b = 0. The first step of the square root method is to make ax^2 = c. We do that by adding 169 to both sides.

4x^2 = 169

We then divide both sides by 4 to isolate x^2.

x^2 = 169/4

To get x by itself we square root both sides. When doing so, the c becomes ±c because both squaring a positive and negative number gets a positive result.

x = ±√(169/4)

We then use the quotient property of square roots to make solving x much easier.

x = ±(√169/√4)

x = ± 13/2

x = { -13/2, 13/2 }

6. 2(x^2) + 5x = -6 (The Quadratic Formula)

The quadratic formula is a derived form of the standard form of a quadratic equation:

x = (-b±√(b^2-4ac))/2a

To use it, we first have to find the a, b, and c in this equation. To do this we must change this into equation into standard form by adding 6 on both sides.

2(x^2) + 5x + 6 = 0

Now we know that a = 2, b = 5, and c = 6. We will now subsitute this into the quadratic formula and solve for x.

x = (-5±√(5^2-4*2*6))/2*2

x = (-5±√(25-48))/4

x = (-5±√(-23))/4

x = (-5/4)±(√(-23)/4)

The square root of -1 is said to be i, an imaginary number. The square root of -23 is √(23)*√(-1) or i√(23).

x = { (-5/4)±(i√(23)/4) }

If don't know imaginary numbers yet or still consider answers with negative square roots as not real, then just say that the equation has no real solutions.

7. x^2 - 6 = 2x (Completing the Square)

Some expressions can be factored so that it multiplies by itself. If we have (a+b)(a+b), it is said to be a perfect square trinomial, where it equals a^2 + 2ab + b^2. The same can be said for (a-b)(a-b), which equals a^2 - 2ab + b^2. Completing the square helps us use this idea to solve quadratic equations.

First, we need to get the quadratic formula into a^2 + bx = c form. So we mush add 6 and subtract 2x on both sides.

x^2 - 2x = 6

Now, in order to make a perfect square trinomial, we consider the b constant. We must add (b/2)^2 to both sides to make sure that we can have this. Only simplify the inside of the parantheses for the first (b/2)^2.

x^2 - 2x + (2/2)^2 = 6 + (2/2)^2

x^2 -2x + (1)^2 = 7

Now that we have a^2 - 2ab + b^2, we can change that to (a-b)^2.

(x-1)^2 = 7

Now look at this! It went from a completing the square problem to a square root method problem. Since we know how to solve this kind of problem, let's do it.

x-1 = ± √7

x = 1 ± √7

x = { 1 - √7, 1 + √7 }

8. z^2 + 81 = 0 (The Square Root Method)

z^2 = -81

z = ±√-81

z = ±(√81*√-1)

z = ±9i

z = { -9i ,9i }

Sadly, there's no example of the factoring method, so I'll make one myself.

BONUS. z^2 - 81 = 0 (Factoring)

First, we make the equation into the form of (x+?)(x+?) = 0.

(x+?)(x+?) = 0.

In the FOIL method, (x+a)(x+b) = x^2 + x(a+b) + ab. What this means is that these two unknown factors add and multiply together to get the b and c constants of the quadratic equation respectively. So, on in the case of b equaling 0 and c equaling -81, we have to ask ourselves this: What two numbers multiply to get -81 and add to get 0?

First, consider the multiplicants: -1 and 81, 1 and -81, -3 and 27, 3 and -27, -9 and 9. Out of all of these, only -9 and 9 fit.

(x+9)(x-9) = 0.

Separate these two into their own equations equal to 0 and solve them.

x + 9 = 0 => x = -9      

x - 9 = 0 => x = 9

x = { -9, 9}