Solving Quadratics, a>1

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

What are the zeros?

• x= 0 and x= -4

• x= 0 and x= 4

• y = 0 and y= 4

• y = 0 and y = -4

Tags

CCSS.HSF-IF.C.7A

True or False:

The solutions, roots, x-intercepts, and zeros of a quadratic equation are all the same thing.

• True

• False, because the solution and root are the same but the x-intercept and zero are different

• False, because the x-intercept and root are the same but the zero and solution are different

• False, they are all different

What are the factors AND solutions of

x² + 2x – 3 = 0

• (x - 2)(x + 1); x = 2, x = -1

• (x + 1)(x - 3); x = -1, x = 3

• (x + 2)(x - 1); x = -2, x = 1

• (x - 1)(x + 3); x = 1, x = -3

Tags

CCSS.HSA-REI.B.4B

Solve

2x² + 17x + 21 = 0

• x= ^-21/₂, -17

• x= ²/₃, 16

• x= -7, ^-3/₂

• x= 15, -³/₇

Which of the following equations matches standard form of a quadratic?

• ax + by = c
• y = a(x - h)² + k
• y = ax² + bx + c
• y = mx + b

Tags

CCSS.HSA.SSE.B.3

GCF of 12 and 18

• 3
• 4
• 6
• 2

Tags

CCSS.7.EE.A.1

Solve:  $3x^2+7x-6=0$ 

•  $x\ =\ 3$  and  $x=-\frac{2}{3}$  

•  $x=-3$  and  $x=\frac{2}{3}$  

•  $x=-3$  and  $x=-\frac{2}{3}$  

•  $x=3$  and  $x=\frac{2}{3}$  

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 multiplication property of zero

• Yes, she factored correctly and then square rooted both sides

• No, the equation must be in standard form set equal to zero before factoring

• No, she didn't factor correctly so the rest of her work is wrong.

Tags

CCSS.HSA-REI.B.4B

What are the factors of

25x² - 81?

• (5x - 9)(5x + 9)

• (5x - 9)(5x - 9)

• (5x + 9)(5x + 9)

• Prime (Can't be factored)

Tags

CCSS.HSA.APR.C.4