I need help with an algebra problem

Algebra
Tutor: None Selected Time limit: 1 Day

√y+5 + √y +13 = 4 (the radical is over y+5 and y+13)  Can someone help me please

Nov 10th, 2015

Thank you for the opportunity to help you with your question!

Hello.  We want to solve for y, correct? 

 sqrt(y+5) + sqrt(y+13)  =  4, is our original equation.

We want to eliminate the radical operations. so we need to square both sides of the equation.

( sqrt(y+5) + sqrt(y+13) )^2 =  4^2

(√(y + 5))^2  + 2*√(y+5) * √(y+13) + (√(y+13))^2  = 16

Expand the left side we get:

where (√(y + 5))^2 = |y + 5|  and  (√(y+13))^2 = |y + 13|

  |y + 5| + 2*√(y+5) * √(y+13) + |y+13| = 16

The right side 4^2  is equal to 16.  Because 4 squared is 16.

We can assume  y > 13  so that  |y + 5| = y + 5 and  |y + 13| = y + 13.

  y + 5 + 2*√(y+5) * √(y+13) + y+13 = 16

 Simplify: 

2y + 18 + 2*√( (y + 5)*(y + 13) ) = 16

2y +2*√ ( (y + 5)*(y + 13) ) = - 2

y + √ ( (y + 5)*(y + 13) ) = -1   this looks like it contradicts the fact that  y > 13, So I have to change my assumption.  

Assume  y < 5  so that  |y + 5| = -(y + 5)   and  |y + 13| = -(y+13)

-(y+5) + 2*√ ( (y + 5)*(y + 13)) - (y + 13) = 16

-y - 5 + 2*√( (y+5)*(y + 13)) - y - 13 = 16

-2y - 18 + 2*√( (y+5)*(y + 13))  = 16

-2y + 2*√( (y+5)*(y + 13)) = 34

-y + √( (y+5)(y + 13)) = 17   from dividing both sides by 2.

√ ( (y+5)(y + 13) ) = y + 17

(y + 5)(y + 13) = (y + 17)^2

y*y + 18y + 16 = y*y + 34y + 289

16 - 289 = 16y

y = -273 / 16



Please let me know if you need any clarification. I'm always happy to answer your questions.
Nov 10th, 2015

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