Solving Quadratics

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

• ±81
• ±9
• ±3
• No real solution

Tags

CCSS.8.EE.A.2

• -4, 2
• -2, 4
• -16
• -4

Tags

CCSS.HSA-REI.B.4B

• -2, 4
• -8
• ±8
• No real solution

Tags

CCSS.HSA-REI.B.4B

If the graph of a quadratic does not intercept the x axis at any point, then it has:

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

Tags

CCSS.HSF-IF.C.7A

Convert x²-6x+9 to factored form and identify the solution.

• (x-3)(x-3); x=3
• (x+3)(x+3); x=-3
• (x+4)(x+2); x=-4, -2
• Not factorable

Tags

CCSS.HSA-REI.B.4B

Find the zeros of this quadratic.

• b=4/5, b=3
• b=4, b=-3
• b=1/5, b=3
• b=4/5, b=-3

Tags

CCSS.HSA-REI.B.4B

Find the zeros.

• 0 and 4
• 0 and -4
• 0
• -4

Tags

CCSS.HSA-SSE.B.3B

CCSS.HSF-IF.C.8A

Determine the values of
a, b, and c for
the quadratic equation: 
4x² – 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

Tags

CCSS.HSA.CED.A.1

CCSS.HSA.REI.B.4

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.B.4B