SOLUTION: Discuss two examples where you use a quadratic function and another where you use a radical equation.

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Question 284186: Discuss two examples where you use a quadratic function and another where you use a radical equation.
Answer by alex224(8) About Me  (Show Source):
You can put this solution on YOUR website!
The standard form of a quadratic function is y = ax^2 + bx + c or f(x) = ax^2 + bx + c where x is the independent variable, y is the dependent variable, and a, b, and c are all constants.

An example of how to solve one of these problems is:

y=x^2-5x-14

1.)Set y equal to zero to find the x-intercepts
0=x^2-5x-14

2.)Next you need to factor
0=(x-7)(x+2)

3.)Then set each factor eqaul to zero
0=x-7 0=x+2

4.)Solve
x=7,-2

Sometimes the function is not factorable and in that case you need to use the quadratic formula:
x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
Here is an example of a problem that needs to be solved by using the quadratic formula:

y=2x^2+4x-5

1.)First you need to set the function equal to zero:
0=2x^2+4x-5

2.)Then you need to determine the variables and plug them into the qudratic equation:
a=2
b=4
c=-5
x+=+%28-4+%2B-+sqrt%28+4%5E2-4%2A2%2A-5+%29%29%2F%282%2A2%29+

x+=+%28-4+%2B-+sqrt%28+16%2B40+%29%29%2F%284%29+

x+=+%28-4+%2B-+sqrt%28+56%29%29%2F%284%29+


4.) Simplify

x+=+%28-1+%2B-+sqrt%28+14%29%29%2F%282%29+

5.)Final Answer

(%28-1+-+sqrt%28+14%29%29%2F%282%29+,0),(+%28-1+%2B+sqrt%28+14%29%29%2F%282%29+,0)

A radical equation is an equation that involves at least one expression or variable under a radical.
To solve these types of equations you need to isolate the variable by doing the opposite operation.

Here is an example:
%28+sqrt%28+x%29%29+=4
In order to get the answer you need to do the opposite of taking the root which is squaring, so you need to sqaure both sides of the equation.

(%28+sqrt%28+x%29%29+)^2=(4)^2

The final answer is x=16