Solving Quadratics

ax²+bx+c=0  is the standard form of a quadratic equation.

• True
• False

Tags

CCSS.HSA.CED.A.1

CCSS.HSA.REI.B.4

Solve the equation: 
5x² - 35x + 60 = 0

• x = 3,  -4
• x = -3,  4
• x = 3,  4
• x = -3,  -4

Tags

CCSS.HSA-REI.B.4B

Solve
x² + 13x + 12 = 0

• x = -12, -1
• x = 12, 1
• x = 13, 1
• x = -13, -1

Tags

CCSS.HSA-REI.B.4B

What is the discriminant?

• b²-4ac
• b²/2a
• 4ac
• b²±4ac

Tags

CCSS.HSA-REI.B.4B

If the discriminant equals 0, then the quadratic has:

• 1 Real Solution
• 2 Real Solutions
• Half a Solution
• No Real Solution

Tags

CCSS.HSA-REI.B.4B

If the discriminant is negative, then the quadratic has:

• 1 Real Solution
• 2 Real Solutions
• Half a Solution
• No Real Solutions

Tags

CCSS.HSA-REI.B.4B

If the discriminant is positive, then the quadratic has:

• 1 Real Solution
• 2 Real Solutions
• Half a Solution
• No Real Solution

Tags

CCSS.HSA-REI.B.4B

What should you do first in solving this equation?
x² + 6x - 13 = 3

• Get factored form
• Write down: a=1, b=6, c=-13
• Make it equal 0 by subtracting 3 on each side
• Type it all in a calculator.

Tags

CCSS.HSA-REI.B.4B

Suzie solves the equation below:
x² + 4x - 12 = 2
(x - 2)(x + 6) = 2, 
by setting x -2 = 0 and x + 6 = 0. 
Her solutions are x = 2 and x = -6.
Is Suzie correct? Why or why not?

• Yes, she factored correctly and then used the zero product property
• Yes, she factored correctly and then square rooted.
• No, she did factor correctly, but she can't use the zero product property as shown.
• No, she didn't factor correctly so the rest of her work is wrong. 

Tags

CCSS.HSA.REI.A.1

CCSS.HSA.REI.B.4

p²+5p+5=0

• -1.11, 1.45
• -0.68, 0.88
• -5.85, 0.85
• -3.62, -1.38

Tags

CCSS.HSA-REI.B.4B