Basics of Algebra
1. Basic Factoring
\[ \text{ Solve for } x: \]
\[ x^2+6x+8 = 0\]
To solve this quadratic equation by factoring, we can look for two numbers that multiply to give 8 and add up to 6. These numbers are 2 and 4.
Therefore, we can rewrite the equation as
\[(x + 2)(x + 4) = 0\]
Setting each factor equal to zero, we have
\[x + 2 = 0 \text{ and } x + 4 = 0\]
\[ \Rightarrow x = -2 \text{ and } x = - 4\]
Hence, the solutions to the equation \(x^2 + 6x + 8 = 0\) are \(x = -2\) and \(x = -4\).
2. Basic Factoring with negative numbers
\[ \text{ Solve for } x: \]
\[ x^2 -6x+8 = 0\]
To solve this quadratic equation by factoring, we are still looking for two numbers that multiply to give 8 but this time they add up to -6. These numbers are -2 and -4.
Therefore, we can rewrite the equation as
\[(x - 2)(x - 4) = 0\]
Setting each factor equal to zero, we have
\[x - 2 = 0 \text{ and } x - 4 = 0\]
\[ \Rightarrow x = 2 \text{ and } x = 4\]
Hence, the solutions to the equation \(x^2 - 6x + 8 = 0\) are \(x = 2\) and \(x = 4\).
2. Basic Factoring with a constant (Part A)
\[ \text{ Solve for } x: \]
\[ 3(x^2 -6x+8) = 0\]
We ignore the 3 and start by factoring the quadratic expression \(x^2 - 6x + 8\). As discussed above, the factored form is \((x - 2)(x - 4)\).
So, we have
\[ 3(x^2 -6x+8) = 3(x - 2)(x - 4)\]
Setting each factor equal to zero, we have
\[x - 2 = 0 \text{ and } x - 4 = 0\]
\[ \Rightarrow x = 2 \text{ and } x = 4\]
Hence, the solutions to the equation \( 3(x^2 - 6x + 8) = 0\) are \(x = 2\) and \(x = 4\).
2. Basic Factoring with a constant (Part B)
\[ \text{ Solve for } x: \]
\[ 4x^2 - 24x + 32 = 0\]
Anytime, we see a number in front of the \( x^2 \) term, we want to see if we can factor it out. In this example, we can first factor out the 4 which makes the equation much easier to solve.
\[ 4x^2 - 24x + 32 = 4(x^2 -6x+8)\]
Now as in the previous problem, we ignore the 4 and start by factoring the quadratic expression \(x^2 - 6x + 8\). As discussed above, the factored form is \((x - 2)(x - 4)\).
So, we have
\[ 4(x^2 -6x+8) = 4(x - 2)(x - 4)\]
Setting each factor equal to zero, we have
\[x - 2 = 0 \text{ and } x - 4 = 0\]
\[ \Rightarrow x = 2 \text{ and } x = 4\]
Hence, the solutions to the equation \( 4x^2 - 24x + 32 = 0\) are \(x = 2\) and \(x = 4\).
2. Factoring out the gcf
\[ \text{ Solve for } x: \]
\[ 4x^3 - 24x^2 + 32x = 0\]
In this advanced version of the previous problem, we aim to factor out common numbers and also variables. By observing the terms, we first factor out the 4 and also the \( x \) which makes the equation much easier to solve.
\[ 4x^3 - 24x^2 + 32x = 4x(x^2 -6x+8)\]
Now as in the previous problems, we ignore the \( 4x \) term and start by factoring the quadratic expression \(x^2 - 6x + 8\). As discussed above, the factored form is \((x - 2)(x - 4)\).
So, we have
\[ 4x(x^2 -6x+8) = 4(x - 2)(x - 4)\]
Setting each factor equal to zero, we have
\[x - 2 = 0 \text{ and } x - 4 = 0
\]
\[ \text{ also } x = 0 \]
\[ \Rightarrow x = 0, x = 2 \text{ and } x = 4\]
Hence, the solutions to the equation \( 4x^3 - 24x^2 + 32x = 0\) are \( x = 0, x = 2\) and \(x = 4\).
2. Two minor modifcations to the quadratic
\[ \text{ Solve for } x: \]
\[ -x^2 + 6x - 8 = 0\]
Any time we see a negative sign in front of the \( x^2 \), we can simplify it by multiplying every term by -1. Now the\( x^2 \) term becomes positive. Then, we can proceed to solve the quadratic equation as usual.
\[ -x^2 + 6x - 8 \rightarrow x^2 - 6x + 8 \]
Now as discussed above, we factor the quadratic as \((x - 2)(x - 4)\).
Setting each factor equal to zero, we have
\[x - 2 = 0 \text{ and } x - 4 = 0\]
\[ \Rightarrow x = 2 \text{ and } x = 4\]
Hence, the solutions to the equation \( -x^2 + 6x - 8 = 0\) are also \(x = 2\) and \(x = 4\).
\[ \text{ Solve for } x: \]
\[ x^2 - 6x = - 8 \]
To solve a quadratic by factoring, we bring all terms to one side and get a 0 on the right hand side. So we add 8 to both sides.
\[ x^2 - 6x = - 8 \] \[ \Rightarrow x^2 - 6x + 8 = 0 \]
Then, we can proceed to solve the quadratic equation as usual.
So, as before, we factor the quadratic as \((x - 2)(x - 4)\).
Setting each factor equal to zero, we have
\[x - 2 = 0 \text{ and } x - 4 = 0\]
\[ \Rightarrow x = 2 \text{ and } x = 4\]
Hence, the solutions to the equation \( -x^2 + 6x - 8 = 0\) are also \(x = 2\) and \(x = 4\).
2. Missing Constant Term
\[ \text{ Solve for } x: \]
\[ x^2 - 6x = 0\]
When the constant term is missing, we simply factor out the \( x \) term
\[ x^2 - 6x = 0 \rightarrow x(x - 6) = 0 \]
Setting each factor equal to zero, we have:
\[ x = 0 \text{ or } x - 6 = 0 \]
Therefore, the solutions to the equation \(x^2 - 6x = 0\) are \(x = 0\) and \(x = 6\).
2. Missing \( x \) Term
\[ \text{ Solve for } x: \]
\[ x^2 - 16 = 0\]
When the \( x \) term is missing, we can start by adding 16 to both sides of the equation.
\[ x^2 - 16 + 16 = 0 + 16\]
\[ \Rightarrow x^2 = 16\]
Now, we can take the square root of both sides to find the solutions.
\[ x = \pm 4\]
Therefore, the solutions to the equation \( x^2 - 16 = 0 \) are \( x = 4 \) and \( x = -4. \)
2. The \(ax^2 + bx + c \) method
All quadratics can be written as \(ax^2 + bx + c = 0\).
When we compare equation \(3x^2 + 5 = 6x\) with \(ax^2 + bx + c = 0\), we get
\[3x^2 + 6x + 5 = 0\]
\[ \Rightarrow a = 3, b = 6, c = 5\]
In the equation \(3x^2 + 5 = 6x\), we first move the \( 6x \) to the left to get the equation \(3x^2 -6x + 5 = 0\). Then we get
\[3x^2 + 5 = 6x\]
\[ \Rightarrow 3x^2 -6x + 5 = 0\]
\[ \Rightarrow a = 3, b =-6, c = 5\]
2. The Quadratic Formula (Your Last Resort)
All quadratics can be solved using this formula:
\[ x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}} \]
Let's solve the quadratic equation below.
\[ 2x^2 - 5x + 3 = 0 \]
\[ \Rightarrow a = 2, b = -5, c = 3\]
Substituting the values \( a = 2, b = -5, c = 3 \) into the formula, we get:
\[ x = \frac{{-(-5) \pm \sqrt{{(-5)^2 - 4(2)(3)}}}}{{2(2)}} \]
Simplifying further, we have:
\[ x = \frac{{5 \pm \sqrt{{25 - 24}}}}{{4}} \]
\[ x = \frac{{5 \pm \sqrt{1}}}{{4}} \]
\[ x = \frac{{5 \pm 1}}{{4}} \]
\[\Rightarrow x = \frac{3}{2} \text{ and } x = 1 \]
Therefore, the solutions to the equation are \( x = \frac{3}{2}\) and \( x = 1\)
6 more sections
Finding the Vertex: Standard Form
y = x^2 - 6x + 8
Finding the Vertex: Intercept Form
y = (x-2)(x-4)
Finding the Vertex: Vertex Form
y = (x-3)^2 + 2
How many real solutions does this equation have?
x^2 -6x + 8 = 0
How many real solutions does this equation have:
x^2 -6x + 9 = 0
How many real solutions does this equation have:
x^2 -6x + 10 = 0
