Quadratic Formula

Quadratic Equations are of the form

 $Ax^2\ +\ Bx\ +\ C\ =\ 0$  

• True

• It's true. This is just a reminder.

• Seriously, take a look at the form that the equation needs to be in to use the quadratic formula, since you're about to use it a bunch of times.

• This is a "poll" question, which means that all the answers are correct. EXCEPT THIS ONE!

Answer explanation

Just kidding - they're all correct.

Determine the values of A,  B, and C for the quadratic equation:

 $4x^2\ -\ 8x\ =\ 3$  

•  $A\ =\ 4,\ B\ =\ -8,\ C\ =\ 3$  

•  $A=4,\ B=-8,\ C=-3$  

•  $A=4,\ B=8,\ C=3$  

•  $A=4,\ B=8,\ C=-3$  

Answer explanation

Don't forget, quadratic equations are of the form  $Ax^2\ +\ Bx\ +\ C\ =\ 0$  , so all of the values must be on the same side of the equation before you can determine A, B, and C.

What would be the best first step in solving this equation?

 $x^2\ +\ 6x\ -\ 13\ =\ 3$  

• Factor

• Write down: a = 1, b = 6, c = -13

• Subtract 3 from both sides.

• Add 3 to both sides.

When a quadratic equation is graphed, it creates a

• Square

• Circle

• Parabola

• Line

Tags

CCSS.HSF-IF.C.7A

If the discriminant equals 0, then the quadratic has:

• 1 Solution

• 2 Real Solutions

• No solutions

• 2 Imaginary Solutions

• 1 Real Solution, and an Imaginary Solution named Harvey.

If the discriminant is negative, you will have....

• two real solutions
• two irrational solutions
• two imaginary solutions
• one solution

Tags

CCSS.HSA-REI.B.4B

Use the graph to determine the solutions.

• -1 and -3
• 1 and -3
• 1 and 3
• -1 and 3

Tags

CCSS.HSA.REI.C.6

CCSS.HSA.REI.C.7

CCSS.HSA.REI.D.10

CCSS.HSA.REI.D.11

Solve using the quadratic formula.

 $2x^2\ -\ 9x\ -35\ =\ 0$  

•  $x\ =\ \frac{7}{2}\$  and  $-6$  

•  $x\ =\ -\frac{5}{2}\$ and  $5$  

•  $x\ =\ \frac{3}{7}\$ and  $6$  

•  $x\ =\ -\frac{5}{2}\$  and  $7$  

Answer explanation

 $\frac{\left(9\pm\sqrt{361}\right)}{4}$  

This graph represents a quadratic equation that will have:

• 2 imaginary roots

• 2 real roots

• 1 real root and 1 imaginary root

Answer explanation

Since the graph doesn't cross the x-axis, no value of x will make the equation equal zero. Thus, the equation has two imaginary roots.

 $2m^2=-2m+12$  
Solve using the quadratic formula.

•  $x=2,-3$  
  

•  $x=3,-2$  
  

•  $x=1\pm\sqrt{13}$  

• No Solution

Answer explanation

 $\frac{\left(-2\ \pm\sqrt{100}\right)}{4}$