Question

Algebra 2 help! D:

Answer 1

Is this correct?

Answer 2

i got x=-4,3 not sure if thats right however

Answer 3

-4.3?

Answer 4

-4 and 3

Answer 5

Ohhh.. well it's only one answer..

Answer 6

First, you can check your answer 5.27 by using that value for x in the equation
you will see you get different numbers on the each side of the equation.
that means 5.27 is not the answer

Answer 7

to solve, square both sides
\[ (x+2)^2 = (\sqrt{3x+16})^2 \\
x^2 +4x+4= 3x+16
\]

Answer 8

I did that part.

Answer 9

you should simplify to get
\[ x^2 +x -12=0 \]

Answer 10

Yes

Answer 11

that factors

Answer 12

Yes with the quadratic function.

Answer 13

you mean quadratic formula. Though the quadratic formula works, you don't need it. factor into
(x+4)(x-3)= 0

that means either x+4=0 or x-3 =0

x=-4 and x= 3
are solutions (as already pointed out by lewopz)

Answer 14

notice that if you test x=3 it works.
also notice that if you test x=-4, it will not work (you get the -2=2 which is not true)
the x=-4 is an extraneous solution

Answer 15

So I would put 3 as the answer..

Answer 16

if you use the quadratic formula you should get
\[ \frac{-1±\sqrt{1+4\cdot 12}}{2} = \frac{-1±\sqrt{1+48}}{2} \\
= \frac{-1±\sqrt{49}}{2}\\
= \frac{-1±7}{2}
= -4 \text{ or } 3
\]

Answer 17

try out your answer x=3

on the left side you get
x+2 becomes 3+2 = 5

on the right side
\[ \sqrt{3x+16}= \sqrt{3\cdot3 +16} = \sqrt{9+16}= \sqrt{25}=5\]

both sides are the same. so 3 works.