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.

(my birthday is September 6, 95 9/6/95)

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.

(my birthday is September 6, 95 9/6/95)

ganeshie8 · Answer

$\large \sqrt{x-y} + m = d$

ganeshie8 · Answer

like that ?

anonymous · Answer

yes! @ganeshie8

ganeshie8 · Answer

good, plugin ur birthday numbers into the equation

anonymous · Answer

ok i did that i'm just lost on finding x

ganeshie8 · Answer

y = year = 85
m = month = 9
d = day = 6

ganeshie8 · Answer

$\large \sqrt{x-y} + m = d$

$\large \sqrt{x-85} + 9 = 6$

ganeshie8 · Answer

subtract 9 both sides

anonymous · Answer

you mean 95 for year right? because thats my year lol

ganeshie8 · Answer

oops ! i made u 10 years old lol ... that should be 95 ah yes..

anonymous · Answer

and 9 from 9 and 6?

ganeshie8 · Answer

yes, 9 from september
6 from day

ganeshie8 · Answer

$\large \sqrt{x-y} + m = d$

$\large \sqrt{x-95} + 9 = 6$

ganeshie8 · Answer

subtract 9 both sides

anonymous · Answer

so sqr root x - 95 = -3
?

ganeshie8 · Answer

$\large \sqrt{x-y} + m = d$

$\large \sqrt{x-95} + 9 = 6$

$\large \sqrt{x-95}  = -3$

ganeshie8 · Answer

since a square root can NEVER be negative. we knw for sure, after solving we will be getting extraneous solution ok ?

ganeshie8 · Answer

square both sides

anonymous · Answer

so $$\sqrt{95} and \sqrt{-3}$$

ganeshie8 · Answer

$\large (\sqrt{x-95})^2 = (-3)^2$

$\large x-95 = 9$

anonymous · Answer

oh alrigth let me calculate that one sec

ganeshie8 · Answer

okie

ganeshie8 · Answer

add 95 both sides ..

anonymous · Answer

i got x = 104

anonymous · Answer

and since it was a -3 it is extraneous right?

ganeshie8 · Answer

x = 104 is correct !

anonymous · Answer

Thank you soooooo much you really broke it down and helped me understand :)

ganeshie8 · Answer

glad to hear that :)
the reasong for it being extraneous is right, but to check whether a solution is extraneous or not, u need to plug it back into the equation and show that it doesnt satisfy the equaiton.

ganeshie8 · Answer

$\large \sqrt{x-95} + 9 = 6$

put x 104 and check if it satisfies above equation or not

anonymous · Answer

Ok thanks again :)

ganeshie8 · Answer

np... u wlc :)