PLEASE HELP!! Algebra 2

No

Edit:

Yes, based on original equation. (Credit to greenpumpkin for correction)

Step-by-step explanation:

For this problem, we simply need to find the values of x that can make the equation true.  So, let's begin by isolating the "x" variable.

sqrt(2x + 13) = x + 5

[sqrt(2x + 13)]^2 = (x + 5)^2

2x + 13 = x^2 + 10x + 25

0 = x^2 + 8x + 12

Note, we can remove the sqrt method by squaring both sides of the equation.  Doing this, we see we have a quadratic equation meaning we can apply the quadratic formula to find solutions for x.

[-b +/- sqrt( b^2 - 4(a)(c) ) ] / 2a

Let a = 1, b = 8, and c = 12

[-8 +/- sqrt( (8)^2 - 4(1)(12) ) ] / 2(1)

= [-8 +/- sqrt( 64 - 48 ) ] / 2

= [-8 +/- sqrt(16) ] / 2

= [ -8 +/- 4 ] / 2

So, x = [ -8 + 4 ] / 2  and x = [-8 - 4 ] / 2

x = [-4] / 2 = -2  and x = [-12] / 2 = -6

Hence, the two values of x that can solve this quadratic equation are x = -2 and x = -6.

Therefore, we know that x = -6 is not extraneous, meaning it is a solution to our equation.

Cheers.

Edit:

Plugging the value of -6 back into the original equation, we get the following:

sqrt(2x + 13) = x + 5

sqrt(2(-6) + 13) = (-6) + 5

sqrt (1) = -1

1 != -1

Given that 1 cannot equal negative 1, we can say that x = -6 is an extraneous solution. (Credit to greenpumpkin for correction)

question 1 is 8, question 2 is 9, question 3 is 24

step-by-step explanation:

now, the numbers they give you at the top are the key. from there, you just have to plug and chug! let's try 1, 2, and 3!

key: a=3, b=5, c=6

1. a+5

remember, a=3, so we plug that in to get 3+5=8

2. 15-c

c=6, so we plug that in to get 15-6=9

3. 4b

b=6, so we plug it in to get 4*6=24

time to try the rest on your own! if you put the answers below in a comment, i'll check them for you!

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step-by-step explanation:

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