Expert: Abe Mantell - 8/6/2006

Question
Hi Abe,

I'm hoping you could help me out with a few of algebra problems I am having difficulty with. I believe these are quadratic equations with roots. I hope I can type it correctly here.


Solve each equation and check for extraneous solutions.

a) the root of w-3 = the root of 4w+15 (I can't make the root sign in this window)


b)the root of x^2-5x+2 = x/2


c) (5/y-3) = 1 + (y+7)/(2y-6)

Can you show all the work that results in the answer? How quickly can you get this back to me?

I have other questions that involve graphing; however, I will send them separately.

Thank you in advance for your expertise. God Bless!


Cindy  

Answer
Hello Cindy,

a) I gather you mean the "square root"...yes?
-- it can be written as: (w-3)^(1/2) = (4w+15)^(1/2)
-- square both sides: w-3 = 4w+15
-- solving for w: -3w = 18 ==> w=-6
-- but -6 cannot be used, assuming you are dealing
-- only with real numbers, since it would give the
-- square root of a negative number.
-- Hence, no solution.

b) (x^2 - 5x + 2)^(1/2) = x/2
-- square both sides: x^2 -5x + 2 = x^2/4
-- multiply both sides by 4: 4x^2 - 20x + 8 = x^2
-- solving for x: 3x^2 - 20x + 8 = 0
-- using the quadratic formula: x=[20 + or - (400-4*4*8)]/8
-- x=(10 + or - 2*sqrt(19))/3

c) I gather you meant: 5/(y-3) = 1 + (y+7)/(2y-6), yes?
-- If so, then multiply by (2y-6): 10 = 2y-6 + y+7
-- solve for y: 9 = 3y ==> y=3, but y cannot equal 3,
-- since it will give a division by zero in the original
-- problem!  So, no solution.

I hope this was helpful!

TTYL, Abe

BTW: What is this for?