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Solving as a quadratic vs. taking the square root

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Solving as a quadratic vs. taking the square root

by katejohn7 » Wed Mar 17, 2010 11:32 am
I am hoping someone can clarify when to solve as a quadratic vs. when to take the square root. For example..

in this question, my instinct when I see a question this "Which of the following could be the value of x - 9 if (x - 2)^2 = 900?"
is to solve as a quadratic, but I am being told I can just take the square root.

This continues to be something I get stuck on and I am sure there is a trusty rule that I have blocked out as a result of all these crazy formulas I am trying to burn into my memory for this test.

Appreciate your feedback.
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by harshavardhanc » Wed Mar 17, 2010 11:41 am
katejohn7 wrote:I am hoping someone can clarify when to solve as a quadratic vs. when to take the square root. For example..

in this question, my instinct when I see a question this "Which of the following could be the value of x - 9 if (x - 2)^2 = 900?"
is to solve as a quadratic, but I am being told I can just take the square root.

This continues to be something I get stuck on and I am sure there is a trusty rule that I have blocked out as a result of all these crazy formulas I am trying to burn into my memory for this test.

Appreciate your feedback.
The point to start is the solid number on the Right hand side (generally) of the equation { variables are there to scare you :) ).


If you see that there is perfect square present there, it will always be easier if you take the square root.


For e.g in the example above, notice that the number is 900 which is a perfect square (30^2)

or if you are able to split the number into a product of perfect squares 9 * 100 ( 3^2 * 10^2).

In these cases, it will be better if you take the square root and tackle a linear equation than handle a second degree equation. :)
Regards,
Harsha
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by harshavardhanc » Wed Mar 17, 2010 11:44 am
but do remember that there will be two cases that you need to consider after taking the square root.

900 will be square of 30 and -30 :)
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Harsha
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by kstv » Wed Mar 17, 2010 8:37 pm
without thinking much
x²=y² take it is obvious x=y but it is valid for x=-y this is less obvious
so if x²= y² x is +y & -y
(x-2)² = 30² so x-2 = 30 and x-2 = -30 x has two values 32 and -28
so we have two values for x-9
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by Stuart@KaplanGMAT » Thu Mar 18, 2010 12:15 pm
harshavardhanc wrote:but do remember that there will be two cases that you need to consider after taking the square root.

900 will be square of 30 and -30 :)
This is the key point to remember and why just taking the root can be dangerous; if you set it up as a quadratic, you'll always see both possible roots.

However, as long as you remember to consider both the negative and positive cases, it's fine to just take the root of both sides (with equations; inequalities are a whole different bag of potential problems).

Another very common mistake that people make is to factor out a variable from both sides. For example, when we have:

x^2 = 10x,

it's very tempting to just divide both sides by x to get:

x = 10;

however, when we do so we're ignoring the possibility that x could also be 0 (in which case we can't simply divide both sides by x).

So, in this case we always want to factor out instead of dividing through:

x^2 = 10x

x^2 - 10x = 0

x(x - 10) = 0

x = 0 or x = 10
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