Verified Answer :
- The Rational Root Theorem (RRT) identifies only rational roots; it fails if any root is irrational or complex.
- \(f(x) = 4x^2 - 25\) has roots \(\pm 2.5\), which are rational, so the RRT list is complete.
- \(g(x) = 4x^2 + 25\) has complex roots \(\pm 1.25i\), so the RRT list is incomplete.
- \(h(x) = 3x^2 - 25\) has irrational roots \(\pm \frac{5\sqrt{3}}{3}\), so the RRT list is incomplete.
- Final conclusions: \(f(x): \text{Yes}\), \(g(x): \text{No}\), \(h(x): \text{No}\).
Explanation
- Understanding the Rational Root Theorem
The Rational Root Theorem (RRT) is a tool for identifying potential rational roots of a polynomial with integer coefficients. For a polynomial \(P(x) = a_n x^n + \dots + a_0\), the theorem states that any rational root \(p/q\) (in simplest form) must have \(p\) as a factor of the constant term \(a_0\) and \(q\) as a factor of the leading coefficient \(a_n\). Crucially, the RRT does not provide information about irrational or complex non-real roots. To determine if the RRT provides a complete list of roots for our functions, we compare the set of all roots (found via the quadratic formula) with the set of candidates provided by the RRT. If any root is irrational or complex, the theorem's list is incomplete.
- Analysis of f(x) = 4x^2 - 25
For \(f(x) = 4x^2 - 25\), the potential rational roots are ratios of the divisors of \(25\) \(\{\pm 1, \pm 5, \pm 25\}\) to the divisors of \(4\) \(\{\pm 1, \pm 2, \pm 4\}\).
Calculating the actual roots:
\[4x^2 - 25 = 0 \implies x^2 = \frac{25}{4} \implies x = \pm \frac{5}{2}.\]
Since \(5/2\) and \(-5/2\) are rational, they are indeed in our set of potential rational candidates. As the quadratic equation has two roots and we found both, the RRT effectively identifies all roots. Thus, for \(f(x)\), the RRT provides a complete list.
- Analysis of g(x) = 4x^2 + 25
For \(g(x) = 4x^2 + 25\), the roots are found by solving:
\[4x^2 + 25 = 0 \implies x^2 = -\frac{25}{4} \implies x = \pm \sqrt{-\frac{25}{4}} = \pm \frac{5}{2}i.\]
The roots are \(\frac{5}{2}i\) and \(-\frac{5}{2}i\). These are non-real complex numbers, not rational numbers. Therefore, the RRT, which only looks for rational roots, cannot provide a complete list of roots for \(g(x)\).
- Analysis of h(x) = 3x^2 - 25
For \(h(x) = 3x^2 - 25\), the roots are found by solving:
\[3x^2 - 25 = 0 \implies x^2 = \frac{25}{3} \implies x = \pm \sqrt{\frac{25}{3}} = \pm \frac{5}{\sqrt{3}} = \pm \frac{5\sqrt{3}}{3}.\]
The roots are \(\frac{5\sqrt{3}}{3}\) and \(-\frac{5\sqrt{3}}{3}\). Since these values are irrational, they cannot be expressed as fractions of integers. The RRT does not list irrational roots, so it fails to provide a complete list of roots for \(h(x)\).
- Conclusion
In summary, the Rational Root Theorem provides a complete list of roots only when all roots of the polynomial are rational. Our analysis shows:
- For \(f(x) = 4x^2 - 25\), the roots are \(\pm 2.5\) (rational), so the RRT list is complete.
- For \(g(x) = 4x^2 + 25\), the roots are \(\pm 1.25i\) (complex), so the RRT list is incomplete.
- For \(h(x) = 3x^2 - 25\), the roots are \(\pm \frac{5\sqrt{3}}{3}\) (irrational), so the RRT list is incomplete.
The final answers are:
\(f(x): \text{Yes}\)
\(g(x): \text{No}\)
\(h(x): \text{No}\)
Examples
The Rational Root Theorem is frequently used in algebra to narrow down the search for roots when you are asked to factorize polynomials or find their x-intercepts. For example, if an engineer is analyzing a physical system modeled by the polynomial \(2x^3 + x^2 - 5x + 2 = 0\), the RRT allows them to quickly test potential values like \(\pm 1, \pm 2, \pm 0.5\), which saves significant time compared to testing arbitrary numbers. While the theorem is powerful, it is important to remember its limitation: it only identifies rational solutions, and other techniques are required to find roots that involve square roots or imaginary units.