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April 13th, 2014, 07:56 PM  #1 
Senior Member Joined: Feb 2014 From: Louisiana Posts: 156 Thanks: 6 Math Focus: algebra and the calculus  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! 
April 13th, 2014, 08:22 PM  #2 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,655 Thanks: 2633 Math Focus: Mainly analysis and algebra 
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 $f(x) \lt 0$. 
April 13th, 2014, 08:50 PM  #3 
Senior Member Joined: Feb 2014 From: Louisiana Posts: 156 Thanks: 6 Math Focus: algebra and the calculus 
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 $\displaystyle x < 2$...

April 13th, 2014, 08:51 PM  #4 
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms  
April 13th, 2014, 08:59 PM  #5 
Senior Member Joined: Apr 2013 Posts: 425 Thanks: 24  
April 13th, 2014, 09:35 PM  #6  
Math Team Joined: Dec 2013 From: Colombia Posts: 7,655 Thanks: 2633 Math Focus: Mainly analysis and algebra  Quote:
$$x + 2 < 0$$ 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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