MHB Why Are There No Real Solutions to the Equation \( |x^2 + 4x| = -12 \)?

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The equation \( |x^2 + 4x| = -12 \) has no real solutions because the absolute value of any expression is always non-negative, meaning \( |x^2 + 4x| \ge 0 \). Since -12 is negative, it is impossible for the left side of the equation to equal -12. Algebraically, both cases for \( x^2 + 4x \) lead to contradictions when checking for real solutions. The discriminant analysis for both cases confirms that there are no valid solutions. Therefore, the equation has no real numbers that satisfy it.
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Explain why there are no real numbers that satisfy the equation
$$|x^2 + 4x| = - 12$$
How is this done algebraically?
 
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RTCNTC said:
Explain why there are no real numbers that satisfy the equation

|x^2 + 4x| = - 12

because ...

$|\text{whatever}| \ge 0$

How is this done algebraically?

solve each case ...

case 1 ... $x^2+4x \ge 0 \implies x^2+4x+12 = 0$ ... recommend looking at the discriminant

case 2 ... $x^2+4x < 0 \implies -x^2-4x+12 = 0$ ... check the two real solutions in the original abs equation.
 

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$|whatever| \ge 0 \implies |whatever| \nless 0$

... that's it.
 
Great. We now move on to the next topic.
 

Thread 'Area of Overlapping Squares'

Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?
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