Strange Solution to Quadratic Equation

[vikcool812]

a strange solution ?

I was solving a tangent problem and came across a strange thing ,
its a simple quadratic equation ,y² - 4y -8 = 0 ,
by quadratic formula , y = 2 + 2*3^1/2 and 2 - 2*3^1/2 ,

now if i write the same quadratic as (y-2)² = 12 ,
then i apply square root function on both sides to get y-2 = (12)^1/2
NOTE : i have not written +,- in front of root 12 because on left side i have a +ve number as it is a square so then i get the answer y = 2*(3)^1/2 + 2
now how can we explain this?

 

[Tac-Tics]

Your mistake is in taking the square root of a square to "undo" it. When x is negative, it is a blatant lie to say \sqrt{x^2} = x. When x is negative, \sqrt{x^2} = -x. Try it out for yourself.

How can you avoid this mistake. Answer: always keep in mind when you root a square. Undoing a square with a root is technically an illegal move unless you know whether x is positive.

How do you legalize this kind of move? You break it into two cases. Draw a vertical line down the middle of your paper. On the left hand side, make the assumption that x is negative. Since every real number must be negative or non-negative, one of the two halves of the paper must logically follow. At the bottom, we get something like y = ... on either side of the page, and those are our solutions.

To illustrate:

y^2 - 4y - 8 = 0
y^2 - 4y + 4 = 12
(y - 2) ^2 = 12

Case 1: Suppose y - 2 \ge 0.

y - 2= \sqrt{12}
y = \sqrt{12} + 2

Case 2: Suppose y - 2 < 0.

y - 2= -\sqrt{12} (note the minus sign on the RHS)
y = 2 -\sqrt{12}

So our two possible solutions are y = \sqrt{12} + 2 or y = 2 - \sqrt{12}.

This is probably the most common mistake every single student of algebra makes.

 

[HallsofIvy]

Or: \sqrt{x^2}= |x|, not just "x".

 

[jambaugh]

It helps to remember that "a solution" does not mean the result of your solving process but a value that makes the original equation true.

This is why we teachers keep harping on "check your answers!". Some mistakes are of the 2+2=5 variety but others are more subtle like this case where you are using a method outside the simpler context in which you originally learned it.

This issue comes up again and again e.g. inverse sine of sine of 7pi/2, and anti-derivatives in calculus. Anytime you are using an inverse function or inverse operation you want to "go on yellow alert" and pay close attention to domain issues.

 

[Tac-Tics]

Anytime you are using an inverse function or inverse operation you want to "go on yellow alert" and pay close attention to domain issues.

Exactly =-) Even for something as common-place as division... remember division doesn't *always* undo multiplication. 2 * 0 / 0 is NOT 2!

 

[HallsofIvy]

It helps to remember that "a solution" does not mean the result of your solving process but a value that makes the original equation true.

I once had a student who complained bitterly that I marked his answer wrong on a test just because he used the "method he learned in high school" rather than the method I taught him. I don't think I was ever able to convince him that I marked him wrong because his answer did not satisfy the equation!