Question

√(4-x)+ √(3x+100) = 0

Answer 1

help me fin its solution set :)

Answer 2

why not get √(4-x) on right hand side and then square both sides?

Answer 3

√(3x+100) = -√(4-x)

3x+100= (-1*√(4-x))^2
3x+100=(-1)^2 * (√(4-x))^2
3x+100= 4-x i think.

Answer 4

then I nedd to find the value of x

Answer 5

yes but now it is more simple :)

Answer 6

ok, thank you for helping me :)

Answer 7

I could be wrong tho :(

Answer 8

its ok :) I appreciate your help :)

Answer 9

√(4-x)+ √(3x+100) = 4
correction :(

Answer 10

its supposed to be four, not zero

Answer 11

oh that makes it harder lol. Imma try

Answer 12

LOL im not good at math T.T

Answer 13

Is this the equation?

\( \sqrt{4-x} + \sqrt{3x+100} = 4\)

Answer 14

yes :)

Answer 15

fidn the solution set :)

Answer 16

You will need to square it twice.
First, you isolate one of the square roots, and you square both sides.
Then you isolate the remaining square root, and you square both sides again.
Finally, when you get answers for x, you check all of them in the original equation because squaring an equation introduces extraneous solution.

Answer 17

okay...

Answer 18

Here we go:

\( \sqrt{4-x} + \sqrt{3x+100} = 4\)

Isolate one root:

\( \sqrt{4-x} = 4 - \sqrt{3x+100}\)

Now when we square both sides, the left root will disappear, but we will still have the one on the right side to deal with.

Answer 19

\( (\sqrt{4-x})^2 = (4 - \sqrt{3x+100})^2 \)

\( 4 - x = 16 - 8\sqrt{3x+100} + 3x + 100 \)

Do you follow so far?

Answer 20

yes :)

Answer 21

where did you get the -8?

Answer 22

The right side was:

\((4 - \sqrt{3x+100})^2 \)

This is the square of a binomial which follows this pattern:

\( (a - b)^2 = a^2 - 2ab + b^2 \)

so \((4 - \sqrt{3x+100})^2 \)

\( = 4^2 - 2(4)(\sqrt{3x+100}) + (\sqrt{3x+100})^2 \)

There you see there the -8 comes from.

Answer 23

ok, I remember that :)

Answer 24

what do we do now?

Answer 25

Let's continue from where we left off.

We got to here:

\(4 - x = 16 - 8\sqrt{3x+100} + 3x + 100 \)

Answer 26

Now we need to again isolate the root, and move everything else to the other side. Then we square both sides.

Answer 27

yessss :)

Answer 28

I added the root to both sides, moving the root to the left side.
Then I subtracted 4 from both sides moving the 4 to the right side.
I added x to both sides moving the x to the right side.

\( 8\sqrt{3x+100} = 16 + 3x + 100 -4 + x\)

Now we combine like terms on the right side:

\( 8\sqrt{3x+100} = 4x + 112\)

Answer 29

Now we have the root n the left side by itself, so we square both sides again:

\( (8\sqrt{3x+100})^2 = (4x + 112)^2 \)

\(64(3x + 100) = (4x)^2 + 2(4x)112 + 112^2 \)

Answer 30

okay :)

Answer 31

\(192x + 6400 = 16x^2 + 896x + 12,544 \)

Answer 32

\(0 = 16x^2 + 704x + 6144 \)

Answer 33

factor it :)

Answer 34

Exactly.
Now we divide both sides by 16 and switch sides:

\(x^2 + 44x + 384 = 0 \)

Answer 35

whooops sorry

Answer 36

Now we need to either factor the quadratic or use the quadratic formula.

Answer 37

its easier if we will use the quadratic formula.

Answer 38

Exactly what I was thinking.

Let's use the quadratic formula.

\( x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a } \)

Here, we have \(a = 1\), \(b = 44\), \(c = 384\).

Answer 39

\( x = \dfrac{-44 \pm \sqrt{44^2 - 4(1)(384)}}{2(1) } \)

\( x = \dfrac{-44 \pm \sqrt{1936 - 1536}}{2 } \)

\( x = \dfrac{-44 \pm \sqrt{400}}{2 } \)

\( x = \dfrac{-44 \pm 20}{2 } \)

\(x = -12\) or \(x = -32\)

Answer 40

Ok, so far?

Answer 41

ok :) so the solution set is the set containing -12 and -32

Answer 42

Not so fast.

Answer 43

yehey! thank you so much for helping me :)

Answer 44

??????

Answer 45

Way above, when I told you the steps you need to follow to solve this equation I wrote:

"You will need to square it twice.
First, you isolate one of the square roots, and you square both sides.
Then you isolate the remaining square root, and you square both sides again.
Finally, when you get answers for x, you check all of them in the original equation because squaring an equation introduces extraneous solution."

Answer 46

Read the last instruction, starting with "Finally, ..."

Answer 47

ok, so its an extraneous solution

Answer 48

I didn't say that. I just said we need to test both soltions we got in the original equation to see if they are solutions or not.

Answer 49

ah okay :)

Answer 50

its the solution

Answer 51

4+8=12 :)

Answer 52

1. Let's test x = -12

\(\sqrt{4-(-12)} + \sqrt{3(-12)+100} =?~ 4\)

\(\sqrt{16)} + \sqrt{64} =?~ 4\)

\(4 + 8 =? ~ 4 \)

\(12 \ne 4 \)

x = -12 does not work.

Answer 53

? im wrong :(

Answer 54

1. Let's test x = -32

\(\sqrt{4-(-32)} + \sqrt{3(-32)+100} =?~ 4\)

\(\sqrt{36)} + \sqrt{4} =?~ 4\)

\(6 + 2 =? ~ 4 \)

\(8 \ne 4 \)

x = -32 does not work.

Answer 55

OMG

Answer 56

Neither solution worked, so the answer is there is no solution.
As a set, the answer is { }, the empty set.

Answer 57

omg, its empty set LOL sorry

Answer 58

I failed in my math exam because I was careless :(

Answer 59

Don't be sorry.
This was a good example of how squaring both sides of an equation can introduce extraneous solutions.
The solutions we got were correct solutions to the quadratic equation, but the quadratic equation did not exactly correspond to the original equation. That's why solutions to the quadratic equation were not solutions to the original equation.

Answer 60

I made plenty of careless mistakes myself. You just need to check your work.

Answer 61

extraneous...big word :) OK, I will remember that :)

Answer 62

thank you :) hahaha

Answer 63

im a neophyte here, how can I give you a medal? :)

Answer 64

Make sure you remember what it means. In our case, the extraneous solutions were false solutions that were introduced by squaring both sides. That's why it's important to check any possible solutions in the original equation.

Answer 65

okay :) my professor in math mentioned that

Answer 66

Click the blue "Best Reposne" box in one of my responses.

Answer 67

done? hahaha thanks :)

Answer 68

wlcm