Problem 112 - Algebra for College Students

Question

For each equation, find approximate solutions rounded to two decimal places. $$1.2 x^{2}-\pi x+\sqrt{2}=0$$

Step-by-Step Solution

Answer

2.04 and 0.58

Step 1: Identify the coefficients

In the quadratic equation $$1.2 x^{2}-\text{π} x+\text{√2}=0$$, identify the coefficients: a = 1.2, b = -π, and c = √2.

Step 2: Use the quadratic formula

The solutions to the quadratic equation ax² + bx + c = 0 can be found using the quadratic formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ Apply the values a = 1.2, b = -π, and c = √2.

Step 3: Simplify under the square root (Discriminant)

First, calculate the discriminant, which is the part under the square root in the quadratic formula: $$b^2 - 4ac = (- \text{π})^2 - 4(1.2)(\text{√2}) = \text{π}^2 - 4 \times 1.2 \times \text{√2}$$ Evaluate $$\text{π}^2 \text{ = 9.87}$$ and $$4 \times 1.2 \times \text{√2 \text{ = 6.78}}$$ Now, $$ \text{Discriminant} = 9.87 \text{ - 6.78} = 3.09.$$

Step 4: Solve for x

Substitute the discriminant back into the quadratic formula: $$x = \frac{-(- \text{π}) \pm \sqrt{3.09}}{2 \cdot 1.2}$$ Simplify this to: $$x = \frac{ \text{π} \pm \sqrt{3.09}}{2.4}$$ Compute the values: \text{π ≈ 3.14} $$ \text{x1 = }\frac{ (3.14) + \text{√3.09}} {2.4} \text{≈ } \frac {3.14 + 1.76} {2.4} = \frac {4.90} {2.4} \text{ ≈ 2.04}$$ $$ \text{x2 = }\frac{ (3.14) - \text{√3.09}} {2.4} \text{≈ } \frac {3.14 - 1.76} {2.4} = \frac {1.38} {2.4} \text{ ≈ 0.58}$$

Key Concepts

Quadratic Formula

When dealing with quadratic equations like the given example, the quadratic formula becomes a powerful tool. The quadratic formula is expressed as:\br> \br> \[ x = \frac{-b \u00b1 \sqrt{b^2 - 4ac}}{2a} \]\br> \br> Here, \(a\), \(b\), and \(c\) are the coefficients from the quadratic equation of the form \(ax^{2} + bx + c = 0\). The formula provides two solutions because of the 'plus-minus' (\(\u00b1\)) in front of the square root. This method can solve any quadratic equation as long as you identify the coefficients correctly. For our specific example, the quadratic equation is \(1.2x^{2}-\text{π}x+\text{√2}=0\). The coefficients are:

• \(a = 1.2\)
• \(b = -\text{π}\)
• \(c = \text{√2}\)

Discriminant

The discriminant is a specific part of the quadratic formula that helps you determine the nature of the roots without solving the equation completely. It is the part under the square root, represented as \(b^{2} - 4ac\). The discriminant can be calculated as follows: \br> \(\text{Discriminant} = (-\text{π})^{2} - 4(1.2)(\text{√2}) = \text{π}^{2} - 4 \times 1.2 \times \text{√2}\)\br> Plugging in the numbers we find:\br> \( \text{π}^2 = 9.87 \) and \( 4 \times 1.2 \times \text{√2} = 6.78 \). \br> Therefore, \( \text{Discriminant} = 9.87 - 6.78 = 3.09 \).\br> \br> If the discriminant is positive, it means there are two real and distinct solutions. If it is zero, there is exactly one real solution. If negative, there are no real solutions, only complex ones.

Approximate Solutions

Sometimes, the solutions to quadratic equations need to be rounded to a certain number of decimal places, especially when dealing with irrational numbers. For this exercise, we need to round it to two decimal places.

• Substitute the discriminant back into the quadratic formula: \[ x = \frac{ (\text{π}) \pm \sqrt{3.09}}{2.4} \]
• To find the two solutions, we need to handle both the plus and the minus separately:
• For \(x_{1}: \frac{ (3.14) + \text{√3.09}}{2.4} \approx \frac{ 3.14 + 1.76}{2.4} = \frac{ 4.90}{2.4} \approx 2.04\)
• For \(x_{2}: \frac{ (3.14) - \text{√3.09}}{2.4} \approx \frac{ 3.14 - 1.76}{2.4} = \frac{ 1.38}{2.4} \approx 0.58\)

\br> Rounding to two decimal places ensures that the solutions are easier to understand and handle in further calculations. So, the approximate solutions for the equation \(1.2x^{2}-\text{π}x+\text{√2}=0\) are \(x\approx 2.04\) and \(x\approx 0.58\).