Question

sqrt(x+4)*sqrt(x+4) = x + 4 right?

Answer 1

\[\LARGE \implies (\sqrt{x + 4}) \times (\sqrt{x+4}) \implies \sqrt{(x+4)\times (x+4)}\] \[\LARGE\implies \sqrt{(x+4)^2} \implies x +4\]

Answer 2

better @Leaper ?

Answer 3

I don't understand why it got cutt off

Answer 4

do you get the process though?

Answer 5

Igba
Do u have Twiddla

Answer 6

no..

Answer 7

\[\sqrt{(x+4)} \times \sqrt{(x+4)}=(x+4)^{\frac{1}{2}} \times (x+4)^{\frac{1}{2}}\]\[ = (x+4)^{\frac{1}{2}+\frac{1}{2}} = (x+4)^1 = x+4\]

Answer 8

\[ \sqrt{x + 4} \times \sqrt{x+4} =\sqrt{16+8 x+x^2}=\sqrt{(4+x)^2} = \pm(x+4) \]

Answer 9

Is sqrt(x^2 + 8x + 16) = x + 4

Answer 10

\[\sqrt{x+4} \cdot \sqrt{x+4}=(x+4)=x+4\]
but
\[\sqrt{(x+4)^2}=\pm (x+4)\]

Answer 11

or \[\huge \sqrt{(x+4)^2} \implies (x+4)^{\frac{2}{2}} = (x+4)^1 \implies x + 4\]

Answer 12

ffm i disagree with what yu have

Answer 13

\[ \sqrt{x + 4} \times \sqrt{x+4} =\sqrt{16+8 x+x^2}\]\[=\sqrt{(4+x)^2} = \pm(x+4) \]

Answer 14

\[\sqrt{x+4} \ge 0\]
it cannot be negative

Answer 15

\[\sqrt{x+4} \cdot \sqrt{x+4} =(x+4)\]

Answer 16

but if we have
\[\sqrt{(x+4)^2}=\pm (x+4)\]

Answer 17

\[\sqrt{x+4} \cdot \sqrt{x+4} \neq \pm (x+4) \]

Answer 18

I don't understand why sqrt(x^2 + 8x + 16) = x + 4

Answer 19

it just equals the positive output

Answer 20

Do you know how to factor x^2+8x+16?

Answer 21

Yes I do

Answer 22

what do you get

Answer 23

If it is \( y^2 = (x+4)^2\) , then \(y = \pm (x+4)\)
but the sqrt there is already positive, why should we make it negative?

If it is \(\pm \sqrt{(x+4)^2}\), then it is \(\pm (x+4)\)
Hmm...

Answer 24

You get (x+4)^2

Answer 25

\[\sqrt{(x+4)^2}=|x+4|=\pm(x+4)\]

Answer 26

By definition of course :)

Answer 27

why is there a minus @myininaya ? doesnt the square root just cancel out thus canceling the \(\pm\) tOO?

Answer 28

I just don't understand this concept, I don't know how to view sqrt(x+2)*sqrt(x+2) = x+2

Answer 29

*totally new to me*

Then FoolForMath is correct?!

Answer 30

\[\LARGE \sqrt a \times \sqrt a = a \longleftarrow \text{rule}\]

Answer 31

\[{\sqrt{x+4}*\sqrt{x+4}=\sqrt{(x+4})^2}\]
\[\huge{x+4}\]

Answer 32

just accept that fact for now @Leaper no need to understand the philosophies as to how, why, etc ;)

Answer 33

i think \(\pm{x+4}\) is not possible

Answer 34

Ok if we have
\[\sqrt{a} \sqrt{a}\]
Then the answer is a
If we have \[\sqrt{a^2}\]
Then the answer is \[\sqrt{a^2}=|a|=\pm a\]

Answer 35

according to the rule as lgba said

Answer 36

of course assuming a>0 :)

Answer 37

for that first one i wrote

Answer 38

What if I have sqrt(10x+50y+20z+30a)*sqrt(10x+50y+20z+30a) does it equal 10x+50y+20z+30a

Answer 39

yep

Answer 40

@myininaya will it be \(\sqrt{a}^2 \) instead of\(\sqrt{a^2}\)

Answer 41

isnt \((\sqrt x)^2 = \sqrt{x^2}\) lol

Answer 42

which one are you talking about @mathslover

Answer 43

no @lgbasallote

Answer 44

Ok if we have
a√a√

Then the answer is a
If we have
a2−−√

Then the answer is
a2−−√=|a|=±a

for this comment

Answer 45

Igba the rule makes sense when sqrt(3x)*sqrt(3x) = 3x because sqrt(3*3*x*x) = sqrt(9x^2) = 3x

Answer 46

yup..you see now? :)

Answer 47

I understand when there is one term not multiple terms under a radical

Answer 48

Whatever I wil accept the rule and try to figure it out later

Answer 49

haha that's right!! i accepted that rule and never complained =))))

Answer 50

which one is the problem what rule?

Answer 51

I'm trying to understand it on a deeper level

Answer 52

\[(\sqrt{a})^2=\sqrt{a} \cdot \sqrt{a}=a , a>0\]
\[\sqrt{a^2}=|a|=\pm(a)\]

Answer 53

the rule that \[\sqrt a \sqrt a = a\] what astounds us though is this new revelation that \(\sqrt{a^2} = \pm a\)

Answer 54

I will tell you my perspective
sqrt(a) = a^1/2, sqrt(a) = a^1/2
When you multiply exponentsn you add the exponents, 1/2 + 1/2 = 2/2 = 1 exponent

a^1/1 = a right?

Answer 55

http://www.wolframalpha.com/input/?i=sqrt%7Bx%2B4%7D*sqrt%7Bx%2B4%7D

Answer 56

but when sqrt(x+5) = x^1/2 + 5^1/2

Answer 57

no no you can't do that leaper

Answer 58

sqrt (x+5) is NOT equal to x^1/2 + 5^1/2

Answer 59

Why

Answer 60

\[\sqrt{a+b}=\sqrt{a}+\sqrt{b}\] that is not correct

Answer 61

imagine this

\[\sqrt{9+ 16}\]

that is originally equal to \(\sqrt{25} = 5\)

but if you try your method you come up with \[\large \sqrt{3^2 + 4^2} = \sqrt{3^2} + \sqrt{4^2} = 3 + 4 = 7\] and that is NOT equal to 5

Answer 62

you get why you cant do it?

Answer 63

let \(\sqrt{a+b}=x\) square both sides ..\(a+b=x^2\)
according to you \(\sqrt{a}+\sqrt{b}=x\) again square both sides ...
\(a+b+2\sqrt{ab}=x^2\) that is not equal to a+b=x^2
hence ur statement is proved as wrong

Answer 64

@lgbasallote has put constants there in the place of a and b that acted as a good example for you hope you now understood ?

Answer 65

@lgbasallote
I will say more about that rule:
\[\sqrt{x^2}=\pm (x)\]
\[\sqrt{x^2}=-x, x<0\]
\[\sqrt{x^2}=x , x>0\]
\[\sqrt{x^2}=0, x=0\]
So Lets test this out
I said if x<0 then sqrt(x^2)=-x
So what is a number less than 0? How about -4
Does that part of the rule hold?
\[\sqrt{(-4)^2}=-(-4)\]
yes :)
We get a positive number because -(-4)=4
:)

Answer 66

Igba so all calculations inside a radical have to be done first before square rooting take action?

Answer 67

yup

Answer 68

yahooo !!!

Answer 69

i am very happy today .... my all LaTeX went crashed..

Answer 70

I think that's that info I was looking for

Answer 71

oh goodie :DDD i was prepared for more examples but if you get it then it's all yahoo

Answer 72

Igba so in sqrt(x+4), 4 has to be added to x before the sum of those two gets square rooted right?

Answer 73

yup...assuming you have a value of x of course

Answer 74

if there's no numerical value to x then you cant add it to 4

Answer 75

Here is another example showing the statement
\[\sqrt{x+y}=\sqrt{x}+\sqrt{y}\]
is false:

\[\sqrt{2}=\sqrt{1+1} \neq \sqrt{1} + \sqrt{1}=1+1=2\]

Answer 76

Thank guys.

Answer 77

Igba regardless if there is a numerical value the x+4 inside sqrt has to be done before square rooting right

Answer 78

like hypothetically speaking

Answer 79

yepp

Answer 80

there is a also a gud question that why is \(\sqrt{-1}=i\) andnot \(\pm i\)

Answer 81

i think this is very true was also told by my maths teacher once but when i asked the teacher he went away ..
as a function it is always + sqrt(25) is +5

but in an equation like x^2 = 5, you will have two solutions +sqrt(5) and - sqrt(5)

Answer 82

Okay finally I understand the problem, the correct way to solve this one is \[ \sqrt{x + 4} \times \sqrt{x+4} =\sqrt{16+8 x+x^2} \] \[ =\sqrt{(4+x)}^2 \color{Red}{\neq \sqrt{(4+x)^2} } = (x+4)\]

Answer 83

we should clear this doubt to the maths inventor ...

Answer 84

More importantly \[ \sqrt{x} \times \sqrt{x}=(\sqrt{x})^2\neq\sqrt{(x)^2} \forall x \in \mathbb{R}\]

But then the OP didn't defined \(x\), we are all assuming it to be real.

Answer 85

And yes roting should be done afterwards.