Problem 47 - Algebra and Trigonometry

Question

\(5-60\) Find all real solutions of the equation. $$ 4(x+1)^{1 / 2}-5(x+1)^{3 / 2}+(x+1)^{5 / 2}=0 $$

Step-by-Step Solution

Answer

Real solutions for \( x \) are \( x = -1, 0, 3 \).

Step 1: Substitute Variables

Let the substitution \( u = (x+1)^{1/2} \). Then, \( u^2 = x+1 \). The equation becomes: \[ 4u - 5u^3 + u^5 = 0 \].

Step 2: Factor the Equation

Factor out the greatest common factor \( u \) from the equation: \[ u(4 - 5u^2 + u^4) = 0 \]. This gives two cases to solve: \( u = 0 \) and \( 4 - 5u^2 + u^4 = 0 \).

Step 3: Solve the Case \( u = 0 \)

If \( u = 0 \), then \((x+1)^{1/2} = 0\), which implies \(x+1 = 0\). Therefore, \(x = -1\).

Step 4: Solve the Polynomial \( 4 - 5u^2 + u^4 = 0 \)

Consider \( v = u^2 \), then the equation becomes \( v^2 - 5v + 4 = 0 \). Solve for \( v \) by factoring: \((v-4)(v-1) = 0\). The solutions are \( v = 4 \) and \( v = 1 \).

Step 5: Back-Substitution for \( v = u^2 \)

Substitute back \( v = u^2 \). For \( v = 4 \), \( u^2 = 4 \) gives \( u = 2 \) or \( u = -2 \). For \( v = 1 \), \( u^2 = 1 \) gives \( u = 1 \) or \( u = -1 \).

Step 6: Final Solution for \( x \) Values

For each \( u \):- If \( u = 2 \), then \( (x+1)^{1/2} = 2 \), so \( x+1 = 4 \) and \( x = 3 \).- If \( u = -2 \), it is not valid since square roots are non-negative.- If \( u = 1 \), then \( (x+1)^{1/2} = 1 \), so \( x+1 = 1 \) and \( x = 0 \).- If \( u = -1 \), it is also not valid for the same reasoning as above.

Key Concepts

Polynomial Equations

Polynomial equations are expressions set to zero, consisting of variables raised to different powers and combined using addition, subtraction, or multiplication. In our exercise example, the equation is \[ 4(x+1)^{1/2} - 5(x+1)^{3/2} + (x+1)^{5/2} = 0 \]which is a polynomial degree equation involving fractional exponents.

Understanding polynomial equations involves recognizing the structure:

• They have terms comprised of coefficients and variables raised to exponents.
• The degree of the polynomial is defined by the highest exponent present after simplification.
• Solving polynomial equations is about finding the zeroes or roots of the equation.

These solutions can be real or imaginary, but for this exercise, we focus on real solutions.

Substitution Method

The substitution method is a handy tool for simplifying complex equations. It involves substituting a part of the equation with a single variable to make the process of solving easier.

In the given equation, we used substitution by letting \( u = (x+1)^{1/2} \). This converts the original equation into a simpler polynomial form \[ 4u - 5u^3 + u^5 = 0 \]. This new equation is easier to manage as it transforms terms with higher root fractions into more familiar polynomial terms.

This concept is especially useful in algebra when dealing with radicals or more complex expressions because:

• It simplifies calculations by reducing complex terms.
• It often results in a straightforward polynomial that can be easily factored or manipulated.
• Once the simpler equation is solved, you can substitute back to find the solutions for the original variable.

Factoring Polynomials

Factoring is a crucial method to solve polynomial equations, especially when derivatives or numeric solutions are too complex. It involves expressing the polynomial as a product of its factors.

In our exercise, after substitution, the polynomial \[ 4u - 5u^3 + u^5 = 0 \]was simplified by factoring out the greatest common factor \( u \), resulting in \[ u(4 - 5u^2 + u^4) = 0 \]. This breaks down the problem by giving potential solutions wherever these factors equal zero.

The benefits of factoring include:

• It helps find solutions more intuitively by finding values that satisfy each factor equal to zero.
• This method ensures that any root solutions are found with fewer steps.
• Factoring is efficient for polynomials with integer or simple fractional roots.

Furthermore, further breaking down the polynomial factor using substitution simplifies it even more, such as introducing \( v = u^2 \), and solving by factoring \((v-4)(v-1)=0\).

Square Root

The square root operation is integral when dealing with equations involving variables raised to a fractional power. It is used to find a value that, when multiplied by itself, gives the original number.

Our original equation uses expressions like \((x+1)^{1/2}\), representing the square root of \(x+1\). Let's explore some main points:

• The square root function is defined only for non-negative values in real numbers.
• When introducing a substitution example \( u = (x+1)^{1/2} \) , care must be taken that even when solved for solutions, results giving square root of negative numbers are invalid.
• The square root assists in simplifying equations into solvable polynomial equations, but negative roots must be handled properly.

Through this exercise, the square roots ensure that each solution remains valid within the domain of real numbers without violating mathematical conventions.