How do you know if a quadratic equation will have one, two, or no solutions?
To determine the number of solutions to an equation of the form:
ax2 + bx + c = 0
you must calculate the value of the discriminant.
The discriminant is equal to b2 – 4ac.
If the discriminant is > 0, then the equation has two real number solutions.
If the discriminant is = 0, then the equation has one real number solution.
If the discriminant is < 0, then the equation has no real number solutions.
How do you find a quadratic equation if you are only given the solution?
Given the solutions “m” and “n”, you can find one possible quadratic
equation by substituting the solutions into the equation:
(x – m)(x – n) = 0
Is it possible to have different quadratic equations with the same solution? Explain.
Yes. Multiplying a quadratic equation by a constant will produce a new
equation with the same solutions as the original equation.
For example, given solutions of 1, and -1, one quadratic equation is:
(x – 1)(x + 1) = 0
x2 – 1 = 0
Multiplying this equation by 5 gives:
5x2 – 5 = 0
which has the same solutions.
Provide your classmates with one or two solutions with which they must create a
quadratic equation.
Solutions: 4, 5
The equation is then:
(x – 5)(x – 4) = 0
x2 – 5x – 4x + 20 = 0
x2 – 9x + 20 = 0
Describe a step-by-step process to solve radical equations. List each step and provide an
explanation of how to complete each step with an example.
To solve a radical equation, follow these steps:
Step 1: Isolate the radical on one side of the equation.
Step 2: Square both sides of the equation.
Step 3: Solve for the variable.
Step 4. Substitute the solutions in the original equation and check for extraneous
solutions.
Here is an example:
Solve :
4x + 5 + 4 = 7
Step 1: Isolate the radical on one side of the equation:
!
4x + 5 + 4 " 4 = 7 " 4
4x + 5 = 3
Step 2: Square both sides of the equation:
!
(
4x +5
)
2
4x + 5 = 9
!
= ( 3)
2
Step 3: Solve for the variable:
4x + 5 = 9
4x + 5 " 5 = 9 " 5
4x = 4
x =1
!
Step 4: Check the solution in the original equation:
4x + 5 + 4 = 7
4 (1) + 5 + 4 = 7
4 +5 +4 = 7
9+4 =7
3+ 4 = 7
7=7
Solution checks.
Last provide a problem to solve.
!
3x +16 "18 = 4
Solution : x = 156
!