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Chapter 2: Problem 37

Explain what is unusual about the solution set of the inequality. $$x^{2}-6 x+12 \leq 0$$

Short Answer

Expert verified
The unusual aspect of the solution set for the inequality \(x^{2}-6x+12 \leq 0\) is that it is an empty set. This is because the roots of the corresponding quadratic equation are complex, which means the quadratic expression does not touch or cross the x-axis for any real number.

Step by step solution

01

Solve the Corresponding Equation

Solve the quadratic equation \(x^{2}-6x+12 = 0\). Find its roots using the quadratic formula \(x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}\). In this case \(a=1, b=-6, c=12\). Calculate the discriminant, \(D=b^2-4ac\). If the discriminant is positive, the roots are real and unequal. If it is zero, the roots are real and equal. If it is negative, the roots are complex or imaginary.
02

Find the Roots

Substitute \(a=1, b=-6, c=12\) into the formula and evaluate, \(x = \frac{-(-6)\pm\sqrt{(-6)^2-4*1*12}}{2*1}\) which simplifies to \(x = \frac{6\pm\sqrt{36-48}}{2}\). As the discriminant is less than 0, the roots are complex number. This implies the quadratic expression is always positive for all real number of \(x\).
03

Formulate the Solution Set

As the quadratic expression is always positive, its values will be always greater than 0. However, in this inequality, it is required the expression to be less or equal to 0. So, there is no any real number of \(x\) which make this inequality valid, so the solution set will be an empty set.
04

Interpret the Solution Set

The unusual aspect of this inequality's solution set is that it is an empty set, meaning there are no real number solutions which make the inequality true. This implies the graph of the quadratic expression is always above the x-axis and doesn't have any x-intercepts.

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