All Algebra II Resources
Example Questions
Example Question #62 : Finding Roots
Find the roots of the quadratic equation
This quadratic can be solved in several different ways, including by graphing or factoring. You can also use the quadratic formula:
This gives us two answers:
Example Question #63 : Finding Roots
Given the equation , find a possible root.
Use the quadratic equation to determine roots of a parabolic function.
The quadratic formula is:
The equation in the form of .
Substitute the known coefficients.
Simplify the equation.
The fraction can be broken into two.
Either or is a possible root.
The answer is:
Example Question #61 : Solving Quadratic Equations
Find the roots of this quadratic:
None of these
Simplify the equation by dividing by 2:
Find the roots using the quadratic equation:
*Because the discriminant (part under the radical) comes to 0, we know our quadratic will have just one repeated solution.
Example Question #64 : Finding Roots
None of these
Complex solutions
Multiply both sides by x:
Put in standard form:
Factor quadratic:
Example Question #62 : Solving Quadratic Equations
Find the roots:
Evaluate by first writing the quadratic equation.
The given equation is already in standard form. Write the coefficients.
Simplify the radical using perfect squares.
The answer is:
Example Question #67 : Finding Roots
Find the roots, if any:
Write the quadratic equation in order to solve for the roots.
The coefficients of the parabola will be used to substitute as values into the given formula.
Substitute the values into the equation.
Simplify the fraction.
Rewrite using factors of perfect squares.
Replace the term.
The answer is:
Example Question #68 : Finding Roots
A ball is launched straight up from the ground at time , and the height of the ball from the ground at time is described by the function . Find the time at which the ball returns to the ground.
In order to solve this problem, we think about how to translate what is asked of us in the problem into a mathematical notion. The problem asks us to find when the ball returns to the ground. We know that the ball is launched from the ground at time , and if we plug into our original function, we will find that at time , . Therefore, our "ground" is located at , which is also the horizontal axis. This means that if we are looking for when our ball returns to the ground, mathematically we are looking for values of which make . In other words, we are looking for the zeros (or horizontal intercepts) of our function.
To find the zeros, we first need to factor the function. The steps of this process can be seen below.
So, we can see that we have two factors - and . Now, as usual when solving for zeros, we need to set the right side of our function equal to 0 and solve for . Because if either of these factors is zero, the entire function will equal zero (as anything multiplied by zero is zero), we can consider each factor separately when set to zero.
Thus, we can see that our two zeros are . Since we know that is when the ball was first launched, we know that when the ball reaches the ground again must be . We can check this by plugging back into the original function and ensuring that it evaluates to zero.
Example Question #69 : Finding Roots
Evaluate the roots for:
Write the quadratic formula.
The coefficients of the variables can be determined by the given equation in standard form: .
Substitute the terms into the equation.
Simplify the terms.
The answer is:
Example Question #70 : Finding Roots
What is a possible root to the quadratic equation?
Write the quadratic equation.
The given polynomial is already in standard form.
Substitute the values into the quadratic formula.
Simplify the top and bottom.
The roots will be complex since we have a negative term inside the radical.
A possible root is:
Example Question #1 : Completing The Square
Complete the square in order to find the vertex of this parabola.
To find the vertex of the parabola, you have to get it into vertex form:
The vertex can then be found at the coordinate .
To get to vertex form, we have to complete the square.
Move the 7 over to the other side by subtracting 7 from both sides of the equation:
You're going to have to add something to both sides of the equation...
...the question now is what. What number, when put in the box, would create a "perfect square" on the right-hand side of the equation?
Well, a perfect square trinomial is one whose factors are the same, like so:
In other words, we're looking for .
Well, if is what goes in the box, and is just , then must equal . Now we can solve for .
And since goes in the box, we need to add 4 to both sides:
Now we can factor the right-hand side very neatly:
After we clean up a bit...
...we get:
That gives us a vertex of .