+2440 I think that the easiest way to prove that \((3n)^2+(4n)^2=(5n)^2\) has infinitely many solutions is to solve for n:
| \((3n)^2+(4n)^2=(5n)^2\) | Just solve for n. |
| \(9n^2+16n^2=25n^2\) | Combine like terms on the left hand side of the equation. |
| \(25n^2=25n^2\) | Notice how both sides of the equation are equal. |
| \(0=0\) | \(0=0\) is a true statement, so every value for n results in a true statement. Since infinitely many solutions exist, more than 2005 solutions exist, and we have solved the problem. |
+2440 If \(f(x)=\sqrt{x-3}\), then, in order to evaluate \(f(12)\), replace all instances of x with 12.
| \(f(x)=\sqrt{x-3}\) | This is the original definition of the function. |
| \(f(12)=\sqrt{12-3}\) | Simplify. |
| \(f(12)=\sqrt{9}\) | |
| \(f(12)=3\) | |
+2440 Hello Guest,
You have been proactive in attempting to simplify the numerator, and you are continuing to make significant headway. However, there is a slight algebraic error lying somewhere in your simplification, but I cannot pinpoint where because of the lack of work provided. Computations like these require extreme perserverance and accuracy. Any small imperfection will result in an answer that is horribly askew.
| \(3(2x^2+9x-5)+(4x^2-1)(x-5)\) | For now, I am worrying about the simplification of the numerator. Let's distribute the 3 into the trinomial. |
| \(6x^2+27x-15+(4x^2-1)(x-5)\) | It is time to expand the product of two binomials. |
| \(\textcolor{red}{6x^2}\textcolor{blue}{+27x}\textcolor{green}{-15}+4x^3\textcolor{red}{-20x^2}\textcolor{blue}{-x}\textcolor{green}{+5}\) | Combine all like terms. I have utilized color coding to show which terms combine. |
| \(\)\(4x^3-14x^2+26x-10\) | |
After this algebra, we are left with \(\frac{4x^3-14x^2+26x-10}{(x+5)(x-5)(2x-1)}\). Notice the discrepancies and try and locate the mistake.
From here, you must factor the numerator and determine whether or not any of the factors of \(4x^3-14x^2+26x-10\) match those of the denominator. Factoring by grouping does not appear to be a valid option here, so you will most likely have to resort to methods such as the rational root theorem and the Descartes' Rules of Signs. To start, though, notice how each term is even, so one can factor out a 2 from every term.
Good luck!
