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.
 
See attachment for my little outline.

View attachment 7492
 

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

... that's it.
 
Great. We now move on to the next topic.
 
Thread 'Erroneously finding discrepancy in transpose rule'

Thread 'Erroneously finding discrepancy in transpose rule'

Obviously, there is something elementary I am missing here. To form the transpose of a matrix, one exchanges rows and columns, so the transpose of a scalar, considered as (or isomorphic to) a one-entry matrix, should stay the same, including if the scalar is a complex number. On the other hand, in the isomorphism between the complex plane and the real plane, a complex number a+bi corresponds to a matrix in the real plane; taking the transpose we get which then corresponds to a-bi...
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