Question

Solve for x, given the equation Square root of x minus 5 + 7 = 11.

Answer 1

\[\sqrt{x}-5+7 = 11 \]

or

\[\sqrt{x-5}+7 = 11?\]

Answer 2

or is it all under sqaure root?

Answer 3

if someone could show mw thw general process or pattern of how to do this equations and others similar that would be helpful.

Answer 4

\[\sqrt{x}-5+7=11\]

Answer 5

Add 5 and subtract 7 to both sides

\[\sqrt{x} = 9\]

Answer 6

81 = x

(9)^2 = 81

Answer 7

x-5 is square rooted

Answer 8

ok

Answer 9

\[\sqrt{x-5}+7 = 11\]

Answer 10

i got x=11.. :(

Answer 11

subtract 7 from both sides

sqrt(x-5) = 4

sqaure both sides

x - 5 = 16
+5 +5

x = 21

Answer 12

how do i know if its extreneous or not ?

Answer 13

wow im really not getting this.

Answer 14

oh wait i just squared both sides. ok

Answer 15

The value under the sqaure root is not negative.

Answer 16

what did you do after you got x=21?

Answer 17

\[\sqrt{x-5}+7=11\]Move the 7 to the right by subtracting it from both sides:
\[\sqrt{x-5}=4\]Square both sides
\[(x-5) = \pm 16\]Solve for both values of \(x\)
\[x-5=+16\]\[x-5=-16\]

Answer 18

Now, here's the crucial part: whenever you solve by squaring a square root like this, you MUST check to see if all of your solutions actually work in the original equation! Many times, one of them will not, and such solutions are called extraneous solutions.

Answer 19

ok but now im at square root thingy over 16 then +7=11. what to do now?

Answer 20

Let's test \(x = 21\)
\[\sqrt{21-5} + 7 = 11\]\[\sqrt{16}+7 = 11\]\[4+7=11\checkmark\]That one is good. However, we also got \[x-5=-16\]\[x = -16+5 = -11\]
\[\sqrt{-11-5}+7 = 11\]\[\sqrt{-16} + 7 = 11\]\[4i + 7 = 11\]That's not true, so that is an extraneous solution.

Answer 21

\[\sqrt{16} = \pm4\]

Answer 22

oh i see

Answer 23

because 4*4 = 16
and
(-4)*(-4) = 16

Answer 24

ok,so how do i get 16 out of square root?

Answer 25

If you didn't know this, you could factor 16: 16 = 2*2*2*2
so \[\sqrt{16} = \sqrt{2*2*2*2} = \sqrt{(2*2)*(2*2)} = 2*2 = 4\]because \[\sqrt{a*a} = a\]so long as \(a\ge0\)

Answer 26

hm ok. i see that.but is 4 going to then be divided into 16?

Answer 27

like where do i go from sqaure root over 16=4? i know 4 goes into it but

Answer 28

there's a convention that the square root sign means the positive or principal square root, so even though \((-4)*(-4) = 16\), \(\sqrt{16} = 4\). When we square both sides, we "forget" whether that was the positive or negative value, so we have to stick in the \(\pm\).

\[\sqrt{16} = 4\]
I'm not sure what you're asking...

Answer 29

im asking, im not sure

Answer 30

so i guess that is the answer

Answer 31

what is the answer?

Answer 32

what do i do if there is a number i front of a square root?

Answer 33

21

Answer 34

i really appreciate this help

Answer 35

\[i = \sqrt{-1}\]so \[i^2 = -1\]

Yes, \(x=21\) is the answer to this problem. We can test it:
\[\sqrt{x-5}+7=11\]\[\sqrt{21-5}+7=11\]\[\sqrt{16}+7=11\]\[4+7=11\]\[11=11\]

Answer 36

Don't be too concerned if the \(i\) stuff confuses you for a while — it took mathematicians several centuries to figure it out the first time :-)

Answer 37

oh ok.

Answer 38

well im really thanking them now.. geeze

Answer 39

in fact, just negative numbers were regarded as highly suspicious for many years!

Answer 40

well what do i do if i have an equation similar to this one but there is a nnumber on the outside of the square root

Answer 41

lol

Answer 42

Not a problem, let's do one!

\[2\sqrt{x-5} + 2 = 10\]

Answer 43

oh goodie

Answer 44

Start the same way, shovel all the loose numbers over to the other side:
\[2\sqrt{x-5}=8\]Now we have a choice, we can divide through by 2, giving us something like we had before, or we can just keep it there. Let's keep it, and square both sides:
\[(2\sqrt{x-5})^2 = 64\]

Answer 45

hmm ok

Answer 46

im follpowiing

Answer 47

we can expand that to \[2*2*(\sqrt{x-5})^2 =\pm 64\]Divide both sides by 4\[(\sqrt{x-5})^2 = \pm16\]

squaring the square root just gives us the stuff under the square root sign (but don't forget about the need for the \(\pm\) and the extraneous solution check)

Answer 48

so that gives us\[x-5 =\pm16\]
and \[x = 21, x = -11\]just like the previous problem :-)

Still, we need to check for extraneous solutions:
\[2\sqrt{21-5}+2 = 10\]\[2\sqrt{16}+2=10\]\[2*4+2=10\checkmark\]
\[x=21\] is a good solution

\[2\sqrt{-11-5}+2 = 10\]\[2\sqrt{-16}+2=10\]\[2*4i+2\ne10\]And just like before \(x=-11\) is an extraneous solution.

Answer 49

Make sense?

Answer 50

ok owd you get x-5-=16

Answer 51

:o

Answer 52

\[(\sqrt{x-5})^2 = \pm16\]

If we square the square root of something, we just get the something:

\[(\sqrt{a})^2 = \sqrt{a}\sqrt{a} = \sqrt{a*a} = a\]

Answer 53

hmm