Question

Substitute your birthday into the equation Square root of the quantity x minus y + m = d, where y is the last two digits of your birth year, m is the month, and d is the day. If you were born on 7/10/1856 like Nikola Tesla, your equation would be Square root of the quantity x minus 56 + 7 = 10. Solve for x and identify if it is an extraneous solution. my7 bday is 3/8/1996

Answer 1

@whpalmer4

Answer 2

@helpme1.2

Answer 3

@radar

Answer 4

I've got an easy one: same month and same day, \(x = \text{year}\) :-)

Answer 5

@radar are you doing your birth year? you must be either much younger or older than you look :-) DI_KAZ' birth year is 96, not 93.

Answer 6

My year is 38 (1938) lol

Answer 7

if you were born a/b/cd, the problem is
\[\sqrt{x-cd} + a = b\]

Answer 8

i got the qution \[\sqrt{x-96}+3=8\]

Answer 9

based on the Tesla example

Answer 10

Right, so how to solve \[\sqrt{x-96}+3=8\]I'd start by moving the 3 to the right hand side. Then you can square both sides and solve for \(x\).

Answer 11

can u show me the steps
im a visaul learner

Answer 12

Okay, subtract 3 from both sides of the equation. What's the new equation?

Answer 13

from which sides the 93 and the 8?

Answer 14

96*

Answer 15

\[\sqrt{x-96} + 3 = 8\]Subtract 3 from both sides of the equation:
\[\sqrt{x-96} + 3 - 3 = 8 -3\]\[\sqrt{x-96}=5\]
Agreed? If we add or subtract the same quantity from both sides, the equation remains equivalent.

Answer 16

RIGHT

Answer 17

Okay, so to get rid of that pesky radical sign on the way to getting \(x=\) all alone, let's square both sides. They are equal, so they should be equal to each other after we square both sides, too.
\[(\sqrt{x-96})^2 = 5^2\]But the square of the square root is just the value under the square root sign:
\[x-96 = 5^2 \]\[x-96=25\]
You can solve that for \(x\), right?

Answer 18

how?

Answer 19

here the ^2 come from

Answer 20

^2 came from squaring each side: I could also have written
\[(\sqrt{x-96})(\sqrt{x-96}) = 5*5\]\[x-96 = 25\]

Answer 21

ok then what

Answer 22

Okay, you've got some money in your pocket. If you hand me 15 bucks, you'll have 10 left. How much money do you have in your pocket?

Answer 23

How much money would you need to have to be able to give me $15 and still have $10 for yourself?

Answer 24

5

Answer 25

If you had $5 in your wallet, you could give me $15 and have $10 left for yourself? I don't think so!

Answer 26

25

Answer 27

If you had $25 in your wallet, you could give me $15, and you would have $25-$15 = $10 left.

That's exactly the form of the problem we are trying to solve.

We've got some number, \(x\), and if we subtract 96 from it, we have 25 left.

Answer 28

\[x-96=25\]If we add 96 to both sides:\[x-96+96 = 25+96\]\[x + 0 = 25 + 96\]\[x=25+96\]\[x =\]

Answer 29

121

Answer 30

yes!

Answer 31

haha thanks

Answer 32

Now we have to test our answer in the original problem to see if it works, or if it is an extraneous solution (read: it doesn't work)

\[\sqrt{121-96} +3 = 8\]\[\sqrt{25}+3=8\]What is the square root of 25?

Answer 33

what number when multiplied by itself gives you 25 as the answer?

Answer 34

5

Answer 35

Right.
\[5+3=8\]True or false?

Answer 36

@DI_KAZ Does the DI stand for Drill Instructor??

Answer 37

true
and yes i want to be one it helps me keep my goals in sight if i see it alot

Answer 38

Good goal.

Answer 39

thanks

Answer 40

You're welcome. whpalmer4 really laid that problem out. Good luck with your studies.

Answer 41

thanks and thanks @whpalmer4

Answer 42

My kid brother was a Marine Harrier pilot during the first Gulf War. Semper Fi!

Answer 43

that awesome !!!

Answer 44

tag me if you ever get stuck on a problem, and I'll try to help you out when I'm next available.