Solving Quadratics - Zero Product Property

Use the Zero Product Property to solve for the values of x.

 $\left(x+2\right)\left(x-3\right)=0$  

• 2 and 3

• -2 and -3

• 2 and -3

• -2 and 3

Solve using the Zero Product Property:
 $\left(3x+2\right)\left(x+1\right)=0$  

•  $x=1,\ x=\frac{2}{3}$  

•  $x=-1,\ \ x=-\frac{2}{3}$  

•  $x=2,\ x=1$  

•  $x=-1,\ \ x=\frac{2}{3}$  

What are the zeros of the function?

  $0=(2x+5)(x-6)$  

•  $x=-6,\ x=\frac{5}{2}$  

•  $x\ =-\frac{5}{2},\ x=6$  

•  $x=-5,\ x\ =\ 6$  

•  $x=5,\ x\ =-6$  

Solve.  

 $r\left(r+7\right)=0$  

•  $r=0,\ r=-7$  

•  $r=0,\ r=7$  

•  $r=-7$  

•  $r=7$  

Solve.

 $0=x(x-6)$  

•  $x=0,\ x=-6$  

•  $x=1,\ x=-6$  

•  $x=0,\ x=6$  

•  $x=6$  

What should you do first in solving this equation?
 $x^2+6x-13=3$  

• Write down:  $a=1x^2,\ b=6x,\ c=-13$  

• Write down:  $a=1,\ b=6,\ c=-13$  

• Make it equal 0 by subtracting 3 on each side

• Make it equal 0 by adding 3 on each side

Suzie solves the equation below:
 $x^2+4x-12=2$ 
 $(x-2)(x+6)=2$  
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

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

• No, she did not set the equation equal to zero before factoring.

Rewrite in factored form and identify the solution.

 $x^2-6x+9=0$  

•  $(x-3)(x-3)=0;\ x=3$  

•  $(x+3)(x+3)=0;\ \ \ x=-3$  

•  $(x+4)(x+5)=0;\ \ \ x=-4\ and\ x=-5$  

• Cannot be factored; no real solution

• A
• B
• C
• D

Tags

CCSS.HSA.APR.B.3

CCSS.HSA.REI.B.4

• A
• B
• C
• D

Tags

CCSS.HSA.APR.B.3

CCSS.HSA.REI.B.4