All Algebra II Resources
Example Questions
Example Question #481 : Intermediate Single Variable Algebra
Find the roots of:
Since the quadratic cannot be factored, there are no roots.
Identify the values of , , and in the standard form of the parabola.
Calculate the discriminant.
Since the discriminant is less than zero, the quadratic is irreducible and there are no real roots. However, there are complex roots. Use the quadratic formula to determine the complex roots.
Example Question #482 : Intermediate Single Variable Algebra
Use the quadratic formula to find the roots of .
and
and
no solution
and
and
The parent function of a quadratic is represented as . The quadratic formula is . In this case , , and . Replacing these values into the quadratic forumula will give you the solutions to the quadratic.
and
Example Question #483 : Intermediate Single Variable Algebra
Solve this quadratic equation by using the quadratic formula:
You must know the quadratic equation .
To plug in the right terms, recognize that polynomials in standard form are symbolized as .
Plug in the values from your equation
simplify within the radical:
Simplify the radical:
Reduce:
Note that this represents two values since there is a in the equation. One is solved with an addition sign and the other is solved with a subtraction sign to yield two answers or roots where this equation crosses the x axis.
Example Question #31 : How To Find The Solution To A Quadratic Equation
Solve for .
1) Begin the problem by factoring the final term. Include the negative when factoring.
–2 + 2 = 0
–4 + 1 = –3
–1 + 4 = 3
All options are exhausted, therefore the problem cannot be solved by factoring, which means that the roots either do not exist or are not rational numbers. We must use the quadratic formula.
Example Question #484 : Intermediate Single Variable Algebra
Solve the quadratic equation:
The standard form of a quadratic equation is , where a,b, and c are constants. Plug these constants into the quadratic formula to solve for x.
Example Question #1621 : Algebra Ii
Solve for by using the quadratic formula:
None of the above
The quadratic equations is:
From here you just plug in your numbers, so:
Simplify:
Then you need to simplify the inside looking for perfect squares:
Each term is divisible by 2, so your final answer is:
Example Question #162 : Solving Quadratic Equations
Use the quadratic formula to solve the equation
, ,
Example Question #29 : Quadratic Formula
Find the zeros of ?
This specific function cannot be factored, so use the quadratic equation:
Our function is in the form where,
Therefore the quadratic equation becomes,
OR
OR
OR
Example Question #30 : Quadratic Formula
Find the roots of .
no real solutions
no real solutions
Use the quadratic equation:
Since the original equation is in standard form, where
.
Therefore,
because the value of the discriminant (the component beneath the square root) is negative, this function has no real solutions.
Example Question #31 : Quadratic Formula
Solve .
No real solutions
No real solutions
This function cannot be factor therefore, use the quadratic equation.
Since the original equation is in the form where
.
Therefore,
Since the value of the discriminant (the value beneath the square root symbol) is negative, this function has no real solutions.