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Certainly! Let's solve the equation [tex]\( |12 - x| = x^2 - 12x \)[/tex]. We need to consider the definition of the absolute value function, which splits the problem into two cases based on the value of [tex]\( x \)[/tex].
### Case 1: [tex]\( 12 - x \geq 0 \)[/tex] (i.e., [tex]\( x \leq 12 \)[/tex])
In this case, [tex]\( |12 - x| = 12 - x \)[/tex]. So, the equation becomes:
[tex]\[ 12 - x = x^2 - 12x \][/tex]
Rearrange the equation to standard quadratic form:
[tex]\[ x^2 - 11x - 12 = 0 \][/tex]
We can solve this quadratic equation using the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = 1 \)[/tex], [tex]\( b = -11 \)[/tex], and [tex]\( c = -12 \)[/tex]:
[tex]\[ x = \frac{11 \pm \sqrt{121 + 48}}{2} = \frac{11 \pm \sqrt{169}}{2} = \frac{11 \pm 13}{2} \][/tex]
This gives us two solutions:
[tex]\[ x = \frac{11 + 13}{2} = 12 \quad \text{and} \quad x = \frac{11 - 13}{2} = -1 \][/tex]
Since [tex]\( x \leq 12 \)[/tex] for this case, both solutions [tex]\( x = 12 \)[/tex] and [tex]\( x = -1 \)[/tex] are valid.
### Case 2: [tex]\( 12 - x < 0 \)[/tex] (i.e., [tex]\( x > 12 \)[/tex])
In this case, [tex]\( |12 - x| = x - 12 \)[/tex]. So the equation becomes:
[tex]\[ x - 12 = x^2 - 12x \][/tex]
Rearrange this equation to standard quadratic form:
[tex]\[ x^2 - 13x + 12 = 0 \][/tex]
Again, we use the quadratic formula, where [tex]\( a = 1 \)[/tex], [tex]\( b = -13 \)[/tex], and [tex]\( c = 12 \)[/tex]:
[tex]\[ x = \frac{13 \pm \sqrt{169 - 48}}{2} = \frac{13 \pm \sqrt{121}}{2} = \frac{13 \pm 11}{2} \][/tex]
This gives us two solutions:
[tex]\[ x = \frac{13 + 11}{2} = 12 \quad \text{and} \quad x = \frac{13 - 11}{2} = 1 \][/tex]
However, only the solution [tex]\( x = 12 \)[/tex] and not [tex]\( x = 1 \)[/tex] satisfies the condition [tex]\( x > 12 \)[/tex]. Therefore, there is no valid solution from this case (since both 12 does not strictly greater than 12).
### Solution Summary:
Combining the results from both cases, we have two solutions: [tex]\( x = 12 \)[/tex] and [tex]\( x = -1 \)[/tex].
Thus, the solutions to the equation [tex]\( |12 - x| = x^2 - 12x \)[/tex] are:
[tex]\[ x = -1 \quad \text{and} \quad x = 12 \][/tex]
### Case 1: [tex]\( 12 - x \geq 0 \)[/tex] (i.e., [tex]\( x \leq 12 \)[/tex])
In this case, [tex]\( |12 - x| = 12 - x \)[/tex]. So, the equation becomes:
[tex]\[ 12 - x = x^2 - 12x \][/tex]
Rearrange the equation to standard quadratic form:
[tex]\[ x^2 - 11x - 12 = 0 \][/tex]
We can solve this quadratic equation using the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = 1 \)[/tex], [tex]\( b = -11 \)[/tex], and [tex]\( c = -12 \)[/tex]:
[tex]\[ x = \frac{11 \pm \sqrt{121 + 48}}{2} = \frac{11 \pm \sqrt{169}}{2} = \frac{11 \pm 13}{2} \][/tex]
This gives us two solutions:
[tex]\[ x = \frac{11 + 13}{2} = 12 \quad \text{and} \quad x = \frac{11 - 13}{2} = -1 \][/tex]
Since [tex]\( x \leq 12 \)[/tex] for this case, both solutions [tex]\( x = 12 \)[/tex] and [tex]\( x = -1 \)[/tex] are valid.
### Case 2: [tex]\( 12 - x < 0 \)[/tex] (i.e., [tex]\( x > 12 \)[/tex])
In this case, [tex]\( |12 - x| = x - 12 \)[/tex]. So the equation becomes:
[tex]\[ x - 12 = x^2 - 12x \][/tex]
Rearrange this equation to standard quadratic form:
[tex]\[ x^2 - 13x + 12 = 0 \][/tex]
Again, we use the quadratic formula, where [tex]\( a = 1 \)[/tex], [tex]\( b = -13 \)[/tex], and [tex]\( c = 12 \)[/tex]:
[tex]\[ x = \frac{13 \pm \sqrt{169 - 48}}{2} = \frac{13 \pm \sqrt{121}}{2} = \frac{13 \pm 11}{2} \][/tex]
This gives us two solutions:
[tex]\[ x = \frac{13 + 11}{2} = 12 \quad \text{and} \quad x = \frac{13 - 11}{2} = 1 \][/tex]
However, only the solution [tex]\( x = 12 \)[/tex] and not [tex]\( x = 1 \)[/tex] satisfies the condition [tex]\( x > 12 \)[/tex]. Therefore, there is no valid solution from this case (since both 12 does not strictly greater than 12).
### Solution Summary:
Combining the results from both cases, we have two solutions: [tex]\( x = 12 \)[/tex] and [tex]\( x = -1 \)[/tex].
Thus, the solutions to the equation [tex]\( |12 - x| = x^2 - 12x \)[/tex] are:
[tex]\[ x = -1 \quad \text{and} \quad x = 12 \][/tex]
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