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April 13th, 2014, 07:56 PM   #1
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Question on Quadratic Inequalities

In the equation:

$\displaystyle x^2 + 4x + 4 < 0$

Why is the solution the empty set?

If:

$\displaystyle x^2 + 4x + 4 < 0$
$\displaystyle (x + 2)^2 < 0$
$\displaystyle x < -2$

Why isn't $\displaystyle x < -2$ the solution? I might be getting something blatantly wrong here, but I would like somebody to clear this up for me. Thanks!
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April 13th, 2014, 08:22 PM   #2
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In the function
$$f(x) = x^2 - (a + b)x + ab = (x - a)(x - b)$$
where $a \le b$, let us see what happens for different values of $x$.

When $x \lt a$, $x - a \lt 0$ and $x -b \lt 0$. Therefore $(x - a)(x - b)$ is the product of two negative numbers and so $f(x) \gt 0$ when $x$ is less than both roots of the quadratic.

When $x = a$, $x -a = 0$. Therefore $(x - a)(x - b)=0$ and so $f(x) = 0$ when $x$ is equal to the lower root.

When $a \lt x \lt b$, $x - a \gt 0$ and $x -b \lt 0$. Therefore $(x - a)(x - b)$ is the product of one positive number and one negative number and so $f(x) \lt 0$ when $x$ is between the roots of the quadratic.

When $x = b$, $x -b = 0$. Therefore $(x - a)(x - b)=0$ and so $f(x) = 0$ when $x$ is equal to the upper root.

When $x \gt b$, $x - a \gt 0$ and $x -b \gt 0$. Therefore $(x - a)(x - b)$ is the product of two positive numbers and so $f(x) \gt 0$ when $x$ is greater than both roots of the quadratic.

Now, in your equation, both roots are equal. This means that there are no values of $x$ that fall between the roots, and therefore, from the analysis above, there are no values of $x$ for which the $f(x) \lt 0$.
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April 13th, 2014, 08:50 PM   #3
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Thank you, your explanation makes perfect sense. However, I am still confused as to why the use of algebraic reasoning in the equation yields the erroneous result of $\displaystyle x < -2$...
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April 13th, 2014, 08:51 PM   #4
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You can't take the square root of both sides of an inequality. Alternately, once you reach
Quote:
Originally Posted by Mr Davis 97 View Post
$\displaystyle (x + 2)^2 < 0$
you know there are no solutions since (real) squares can't be negative.
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April 13th, 2014, 08:59 PM   #5
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Quote:
Originally Posted by Mr Davis 97 View Post
In the equation:$\displaystyle x^2 + 4x + 4 < 0$
The inequality $\displaystyle x^2+4x+4<0$ is equivalent with the equation
$\displaystyle x^2+4x+4=a$ where $\displaystyle a<0$.
How to solve the equation $\displaystyle x^2+4x+4-a=0$ where $\displaystyle a<0$?
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April 13th, 2014, 09:35 PM   #6
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Quote:
Originally Posted by Mr Davis 97 View Post
the equation yields the erroneous result of $\displaystyle x < -2$...
The correct next line would be
$$|x + 2| < 0$$

But an absolute value can't be less than zero.

Or you notice that no square of a real can be less than zero as CRGreathouse noted.

Or you notice that Dacu's equation has no real roots for $a \lt 0$.
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