Part a)
for extraneous solution
<span>
1 ⋅ sqrt(x+2) + 3 = 0</span>
for non extraneous solution
1⋅ sqrt(x+2) + 3 = 6
part b) solve each equation
1⋅sqrt(x+2)+3=0
x+sqrt(x+2)=−3
square both sides
(sqrt(<span>x+2)</span>)^ 2 = (−3)^2
x+2=9
x=9−2
x=7
do you see why its extraneous
Answer:
.46 x 14 = 6.44
46 x 1.4 = 64.4
46 x 14 = 644
4.6 x 14 = 64.4
46 x .14 = 6.44
Step-by-step explanation:
14. (2x - 1)(x + 7) = 0Using the zero factor property, we know that either the first or second terms (or both) must be equal to 0 if their product is 0. We can set each term equal to 0 to find the solutions:
2x - 1 = 0
2x = 1
x = 1/2
x + 7 = 0
x = -7
15. 
To solve this equation, you first need to set it equal to 0:

Next, it can be factored:

Finally, we can solve just like we did above:
x + 5 = 0
x = -5
x - 2 = 0
x = 2
16. 
First, you can simplify by dividing each side by 4:

Now, set the equation equal to 0:

Next, factor:

Finally, find the solutions:
x + 5 = 0
x = -5
x - 5 = 0
x = 5
Answer:
9x = 1-1 is q r x= 2x2 + 2 is 4