The3Mathketeers

I am confused with this answer. Why is it that the interval that satisfies neither inequality is the union of the intervals found? I would expect the interval that satisfies neither of the inequalities to be the intervals of real numbers not contained within the original interval. That would be \((-1, 6)\). This is the interval of real numbers that is greater than -1 and less than 6.

ElectricPavlov's method is perfectly valid, but there is a method that avoids having to do arithmetic with square roots, which can become quite messy. I mostly thank Cphill for presenting Vieta's formula in a previous answer. Make sure the quadratic equation is in standard form first.

\(3x^2 + 4x - 9 = 2x^2 - 6x + 1 \\ x^2 + 10x - 10 = 0\)

By Vieta's formula, we can deduce the product of the roots is the constant term, and the sum of the roots is the opposite of the coefficient of the x-term. This means that \(r_1 + r_2 = -10 \text{ and } r_1r_2 = -10\). Our ultimate goal is to find \(r_1^2 + r_2^2\).

\(r_1 + r_2 = -10 \\ r_1^2 + 2r_1r_2 + r_2^2 = 100 \\ r_1^2 + 2 * -10 + r_2^2 = 100 \\ r_1^2 + r_2^2 = 120\)

We have found the sum of the square of the roots!

First, simplify the quadratic into its standard form. This will allow you to identify information about the roots of the equations.

\(2x^2 + 6x - 14 = x^2 - 8x + 2 \\ x^2 + 14x - 16 = 0\)

We can also expand the product of the two binomials. This will be important shortly, as you will see.

\(\begin{align*} (2a - 3)(4b - 6) &= 8ab - 12a - 12b + 18 \\ &= 8ab - 12(a + b) + 18 \end{align*}\)

We can use Vieta's formula to find the product of the roots and the rum of the roots. For a quadratic, the product of the roots is the constant term, and the sum of the roots is the opposite of the coefficient of the x-term.

\(ab = -16, a + b = -14 \\ \begin{align*} 8ab - 12(a + b) + 18 &= 8 * -16 - 12 * -14 + 18 \\ &= -128 + 168 + 18 \\ &= 58 \end{align*} \)

0.13 is written to the hundredth place, which means that its corresponding fraction can be represented as a fraction with 100 as the denominator.

\(0.13 = \frac{13}{100}\)

Here are a few more examples that may make this concept make sense.

0.7 is written to the tenth place, which means that its corresponding fraction can be represented with 10 as the denominator.

\(0.7 = \frac{7}{10}\)

0.111 is written to the thousandth place, which means that its corresponding fraction can be represented with 1000 as the denominator.

\(0.111 = \frac{111}{1000}\)

I tried to read SpectraSynth's answer, and I was unable to make sense of it. I arrived at a different answer than SpectraSynth, so I will present my method, and you can figure out which answer is the correct answer.

Our ultimate goal is to find the area of Rectangle ABCD. If we knew the length and the width of the rectange, then finding the area would be trivial because \(A_{\text{rect}} = l \times w\) where \(l\) is the length and \(w\) is the width. As such, I will label \(AB = l\) and \(AD = w\). 

By definition, we know that Rectangle ABCD is a quadrilateral with four right angles. This observation lead me to realize that \(\triangle \text{ABE} \text{ and } \triangle \text{ADF} \text{ and } \triangle \text{ECF}\) are all right triangles. Because they are right triangles, finding a relationship between the lengths of the triangle legs and the area will be easier to deduce. \(\overline{BE}\) is the height of \(\triangle \text{ABE}\) and \(\overline{\text{DF}}\) is the height of \(\triangle \text{ADF}\). I will let \(BE = x \text{ and } AD = y\). With this information, I can already generate two equations that relate the triangle legs and the area. I simplified these equations as much as I could.

We can employ a similar strategy for \(\triangle \text{ECF}\). However, it is almost always best to relate new variables to information that you already have. Because figure ABCD is a rectangle, we can relate the \(\text{EC} \text{ and } \text{EF}\) to other lengths we already know.

Now that we have related it to variables we have already defined, we can use the area formula for rigth triangles to find more information.

\(A_{\triangle{\text{ECF}}} = \frac{1}{2} * \text{EC} * \text{CF} \\ 10 = \frac{1}{2}(w - x)(l - y) \\ 20 = lw - wy - lx + xy\)

We can already simplify this complicated equation by recognizing that we know from previous equations that \(lx = 10 \text{ and } wy = 10\).

\(20 = lw - 10 - 10 + xy \\ lw + xy = 40\)

This equation is close to solved because we only need to solve for \(lw\), which is equivalent to the area of Rectangle ABCD. We can use previous equations to substitute in an expression for \(xy\). We can multiply previous equations together to find a relationship in which we can ultimately use substitution.

\(lx = 10; wy = 10 \\ lxwy = 100 \\ xy = \frac{100}{lw}\)

Now, we can substitute this relationship into the equation and solve for \(lw\).

\(lw + xy = 40; lw = \frac{100}{lw} \\ lw + \frac{100}{lw} = 40 \\ (lw)^2 + 100 = 40lw \\ (lw)^2 - 40(lw) + 100 = 0\)

The equation simplifies to a quadratic equation, which I will solve with the trusty quadratic formula.

\({\color{red}1}(lw)^2 {\color{blue} - 40}(lw) {\color{green}+ 100} = 0 \\ lw_{1, 2}=\frac{-({\color{blue}-40}) \pm \sqrt{({\color{blue}-40})^2 - 4 * {\color{red}1} * {\color{green}100}}}{2 * {\color{red}1}} \\ lw_{1, 2} = \frac{40 \pm \sqrt{1600 - 400}}{2} \\ lw_{1, 2} = 20 \pm \frac{20\sqrt{3}}{2} \\ lw = 20 - 10\sqrt{3} \text{ or } lw = 20 + 10\sqrt{3} \\ lw \approx 2.6795 \text{ or } lw \approx 37.3205\)

The quadratic yields two different answers, and they both seem logical. Are there two answers? Upon studying the diagram carefully, I rejected \(lw = 20 - 10\sqrt{3}\) because Rectangle ABCD has a minimum area of \(5 + 5 + 10 = 20\) because the rectangle contains three triangles of which we know the area and one non-right triangle of which we do not know the area. Since \(20 - 10\sqrt{3} < 20\), I rejected this answer.

Therefore, \(A_{\text{Rectangle ABCD}} = lw = 20 + 10\sqrt{3} \approx 37.3205\)

The discriminant is absolutely useful here. The discriminant gives insight about the number of solutions for a particular quadratic equation in standard form. When the discriminant equals zero, then there is one and only one solution.

\(4x^2 + nx + 25 = 2x^2 + 5 \\ 2x^2 + nx + 20 = 0\)

Now, the equation is in standard form, so we can use the information from the discriminant to determine values of n that yield one and only solution in the x-variable.

\({\color{red}2}x^2 + {\color{blue}n}x + {\color{green}20} = 0 \\ \Delta = {\color{blue}b}^2 - 4{\color{red}a}{\color{green}c} \\ \Delta = {\color{blue}n}^2 - 4*{\color{red}2}*{\color{green}20} \\ \Delta = n^2 - 160\)

Set the discriminant to zero to find the n-values that lead to one and only one solution.

\(n^2 - 160 = 0 \\ n^2 = 160 \\ n = \pm \sqrt{160} \\ n = 4\sqrt{10} \text{ or } n = -4\sqrt{10}\)

The question asks for the positive value of n only, so we reject the negative answer. Therefore, \(n = 4 \sqrt{10}\)

This problem involves the addition of rational functions. Since fractions are involved, we have to consider when the denominator equals 0. This process is made easier by the process of factoring.

\(f(x) = \frac{x^2 + 10x + 21}{x^2 - 4x -21} + \frac{x^2 - 1}{x^2 - 4x + 4} \\ f(x) = \frac{x^2 + 10x + 21}{(x + 3)(x - 7)} + \frac{x^2 - 1}{(x - 2)^2}\)

After factoring the denominators, we can identify the x-values that will be outside of the domain.

\(x + 3 \neq 0 \quad x - 7 \neq 0 x - 2 \neq 0 \\ x \neq -3 \quad x\neq 7 \quad x \neq 2\)

All of these x-values are not within the domain.

Therefore, the domain in interval notation is \((-\infty, -3) \cup (-3, 2) \cup (2, 7) \cup (7, \infty)\)

\(f(x)\) is a rational function, so an x-value that could realistically not be in the domain occurs when the denominator equals 0.

\((x^2 - 6x + 8) - (x^2 + x - 6) = 0 \\ x^2 - 6x + 8 - x^2 - x + 6 = 0 \\ -7x+14 = 0 \\ -7x = -14 \\ x = -2\)

The value of x = -2 causes the denominator to be 0, x = -2 is the only x-value not in the domain of this function!