Problem 23 - Algebra 2

Question

Solve each equation using the Quadratic Formula. Find the exact solutions. Then approximate any radical solutions. Round to the nearest hundredth. $$ 3 x^{2}-10 x+5=0 $$

Step-by-Step Solution

Answer

The exact solutions are \(x = \frac{10 \pm \sqrt{40}}{6}\). The approximate solutions to the nearest hundredth are \(x \approx 2.72\) and \(x \approx 0.61\).

Step 1: Identify Coefficients

For the quadratic equation in standard form, ax^2 + bx + c = 0, identify the coefficients a, b, and c. In this case, a = 3, b = -10, and c = 5.

Step 2: Apply the Quadratic Formula

Use the quadratic formula to find the roots of the equation, which is \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). Substitute a = 3, b = -10, and c = 5 into the formula to find the solutions for x.

Step 3: Calculate the Discriminant

Calculate the discriminant, which is the term under the square root in the quadratic formula: \(\Delta = b^2 - 4ac = (-10)^2 - 4(3)(5)\).

Step 4: Simplify the Discriminant

Compute the value of the discriminant: \(\Delta = 100 - 60 = 40\).

Step 5: Evaluate the Square Root

Evaluate the square root of the discriminant: \(\sqrt{\Delta} = \sqrt{40}\).

Step 6: Write the Solutions

Write out both solutions using the quadratic formula and the evaluated discriminant: \(x = \frac{-(-10) \pm \sqrt{40}}{2(3)}\).

Step 7: Simplify Further

Simplify the solutions by performing the arithmetic: \(x = \frac{10 \pm \sqrt{40}}{6}\).

Step 8: Approximate Radical Solutions

Approximate the radical \(\sqrt{40} \approx 6.32\) and simplify both solutions to get approximate values: \(x \approx \frac{10 \pm 6.32}{6}\).

Step 9: Round to the Nearest Hundredth

Round the solutions to the nearest hundredth: \(x \approx \frac{10 + 6.32}{6} \approx 2.72\) and \(x \approx \frac{10 - 6.32}{6} \approx 0.61\).

Key Concepts

Solving Quadratic Equations

Understanding how to solve quadratic equations is crucial for students in algebra. A quadratic equation is a second-degree polynomial typically in the form of \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are coefficients, and \(a\) cannot be zero, because if it were, the equation would no longer be quadratic.

To find the roots (solutions) of such an equation, we can use the Quadratic Formula, which provides an exact solution by accounting for all the coefficients. The formula is \(x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}}\). You first need to identify your \(a\), \(b\), and \(c\) values, which in the example of \(3x^2 - 10x + 5 = 0\) are 3, -10, and 5, respectively. Plugging these coefficients into the formula gives you the precise values for \(x\) that satisfy the equation.

Discriminant in Quadratic Equations

The discriminant is a key element within the Quadratic Formula that decides the nature and number of solutions for the quadratic equation. It is found within the square root of the formula and is represented by \(\Delta = b^2 - 4ac\).

The discriminant can tell us:

• If \(\Delta > 0\), there are two real and distinct solutions.
• If \(\Delta = 0\), there is one real repeated (double) solution.
• If \(\Delta < 0\), there are no real solutions; instead, there are two complex solutions.

Calculating the discriminant for the equation \(3x^2 - 10x + 5 = 0\) gives us \(\Delta = (-10)^2 - 4(3)(5)\), which simplifies to 40, indicating that there are two real and distinct solutions for this quadratic equation.

Approximating Radical Solutions

Once the discriminant is calculated and we know that radical solutions are part of the answer, we can proceed to approximate them. A radical solution involves square roots, and they can sometimes be irrational numbers, meaning they cannot be neatly expressed as a simple fraction.

Real-world applications often need an approximate or rounded value, so we evaluate the square root and then use rounding as necessary. For instance, \(\sqrt{40}\) is not a neat number; it's approximately 6.32. Thus, to continue with the solution for the provided quadratic equation, we use this approximation to find the values of \(x\) that are precise enough for practical purposes. Accurate approximations allow a balance between exact mathematical solutions and their practical applications.

Rounding to the Nearest Hundredth

Rounding is a mathematical technique used to make numbers simpler or to indicate precision. Rounding to the nearest hundredth means adjusting the number to two decimal places. It's crucial when dealing with irrational numbers like square roots or in situations demanding precision to a particular level.

This step is final in solving quadratic equations when we are seeking a neat answer to present, especially if the solution will be used in further calculations or as part of measurements. Using our example, after the approximations, the values for \(x\) are rounded: \(x \approx \frac{10 + 6.32}{6} \approx 2.72\) and \(x \approx \frac{10 - 6.32}{6} \approx 0.61\). Correctly rounding these numbers helps ensure that we maintain the accuracy while also providing a clear and usable result.