Problem 40 - College Algebra Essentials

Question

Solve each equation with rational exponents. Check all proposed solutions. $$ \left(x^{2}-3 x+3\right)^{\frac{3}{2}}-1=0 $$

Step-by-Step Solution

Answer

The solutions to the equation are \( x = 2, x = 1 \)

Step 1: Isolate the term with rational exponent

The first step is to isolate the term with the rational exponent. You can do this by adding one to both sides of the equation. The equation becomes: \[ \left(x^{2}-3 x+3\right)^{\frac{3}{2}} = 1 \]

Step 2: Remove the rational exponent

Next, we want to rid of the \(\frac{3}{2}\) exponent. Do this by raising both sides to the power of \(\frac{2}{3}\). This results in: \[ x^{2}-3x+3 = 1^\frac{2}{3} \] which simplifies to: \[ x^{2}-3x+3 = 1 \]

Step 3: Solve the quadratic equation

Now, we have a quadratic equation. Rearrange the terms to set the equation to zero: \[ x^{2}-3x+2 = 0 \] The next step is to solve this equation either by factoring, completing the square, or using the quadratic formula. Factoring the equation, we get: \[ (x-2)(x-1) = 0 \] Setting each factor equal to zero gives us the roots of the equation: \[ x = 2, x = 1 \]

Key Concepts

Quadratic Equations

Quadratic equations are a type of polynomial equation where the highest power of the variable is two. They are generally written in the form \(ax^2 + bx + c = 0\). Here, \(a\), \(b\), and \(c\) are constants, and \(x\) is the variable we want to solve for. Quadratic equations can have up to two solutions, known as roots.
Representations of quadratic equations in terms of their graphs are parabolic shapes.

• If \(a > 0\), the parabola opens upwards.
• If \(a < 0\), the parabola opens downwards.

Quadratic equations are solved using different methods, like factoring, completing the square, or applying the quadratic formula.
They are very resourceful in mathematics as they can describe real-world phenomena such as projectile motion and areas of land.

Factoring Quadratic Equations

Factoring quadratic equations is a method used to find the roots by expressing the equation as a product of two binomials. To achieve this, one needs to decompose the quadratic equation into two linear factors. The equation \(x^2 - 3x + 2 = 0\) from the exercise can be factored into

• \((x - 2)(x - 1) = 0\)

This approach relies on identifying two numbers that simultaneously add up to the coefficient of \(x\), which is \(-3\), and multiply to the constant term, which is \(2\).
Once factored, solving the equation is straightforward. Set each factor equal to zero and solve for \(x\). Hence, the equation (\(x - 2\)) gives \(x = 2\), and (\(x - 1\)) gives \(x = 1\). These solutions indicate the points where the parabola intersects the x-axis.

Solving Equations with Exponents

Solving equations with exponents can sometimes involve rational exponents, which are exponents that are fractions. Rational exponents provide a way to express roots using powers. For example, in the exercise \(\left(x^{2}-3 x+3\right)^{\frac{3}{2}}-1=0\), the exponent \(\frac{3}{2}\) means take the square root of the expression and then cube the result.
To solve such equations, the initial step typically involves isolating the term with the exponent on one side, as showed in the original solution. After isolation, you raise both sides to the reciprocal power to eliminate the exponent.

Here are some tips for solving such equations:

• Keep the equation balanced by performing the same operation on both sides.
• Always check your solutions, as operations involving exponents can introduce extraneous solutions.
• Use integer powers where possible to simplify the calculations.

Mastering equations with exponents is key, not only for solving these types of equations, but also in broader areas of math and science. They enable you to tackle more complex equations involving polynomial roots and transformations.