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Chapter 8: Problem 26

Solve each equation by using the quadratic formula. $$\frac{3}{4} x^{2}-2 x+\frac{1}{2}=0$$

Short Answer

Expert verified
2.387 and 0.279

Step by step solution

01

Identify coefficients a, b, and c

In the given quadratic equation \(\frac{3}{4} x^2 - 2x + \frac{1}{2} = 0\), the coefficients are: \(a = \frac{3}{4}\), \(b = -2\), \(c = \frac{1}{2}\).
02

Write down the quadratic formula

The quadratic formula to solve for \(x\) is: \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\].
03

Substitute the coefficients into the quadratic formula

Substitute \(a = \frac{3}{4}\), \(b = -2\), and \(c = \frac{1}{2}\) into the quadratic formula: \[x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4 \cdot \frac{3}{4} \cdot \frac{1}{2}}}{2 \cdot \frac{3}{4}}\].
04

Simplify within the square root

Calculate within the square root: \((-2)^2 = 4\) and \(-4 \cdot \frac{3}{4} \cdot \frac{1}{2} = -1.5\). \[x = \frac{2 \pm \sqrt{4 - 1.5}}{1.5}\].
05

Compute the discriminant

Simplify the expression under the square root: \[4 - 1.5 = 2.5\]. So we have: \[x = \frac{2 \pm \sqrt{2.5}}{1.5}\].
06

Solve for the values of x

Solve for \(x\) by computing each part: \[x = \frac{2 + \sqrt{2.5}}{1.5}\] and \[x = \frac{2 - \sqrt{2.5}}{1.5}\].
07

Simplify the solutions

Calculate the numerical values for the solutions: \[x = \frac{2 + 1.581}{1.5} \approx 2.387\] and \[x = \frac{2 - 1.581}{1.5} \approx 0.279\].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

solving quadratic equations
When dealing with quadratic equations, the quadratic formula is a powerful tool. A quadratic equation is of the form: \[ax^2 + bx + c = 0\] The quadratic formula solves for the variable \(x\) and is given by: \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\] To use this formula, identify the coefficients \(a\), \(b\), and \(c\) from the equation. These coefficients are: \[a, b, c\]. Input these values into the formula to find the solutions for \(x\). Remember, the \(\pm\) symbol implies that there are usually two solutions: one by adding the \( \sqrt{b^2 - 4ac }\) and one by subtracting it.
discriminant
The discriminant is a key part of the quadratic formula contained within the square root: \[b^2 - 4ac\]. It determines the nature of the roots of the quadratic equation:
  • If \(b^2 - 4ac > 0\), there are two distinct real roots.
  • If \(b^2 - 4ac = 0\), there is exactly one real root, also called a repeated or double root.
  • If \(b^2 - 4ac < 0\), there are no real roots (the roots are complex or imaginary).
Calculating the discriminant is an important step as it informs you about the number and type of solutions you will get when solving the quadratic equation.
coefficients
Coefficients in a quadratic equation, denoted as \(a\), \(b\), and \(c\), are the constants that multiply the variable terms. In the generic form of a quadratic equation: \[ax^2 + bx + c = 0\]
  • \(a\) is the coefficient of \(x^2\). It affects the width of the parabola when the quadratic is graphed.
  • \(b\) is the coefficient of \(x\). It influences the location of the vertex along the x-axis.
  • \(c\) is the constant term. It moves the parabola up or down on the graph.
Identifying these coefficients correctly is crucial when substituting them into the quadratic formula. In our example, \(a = \frac{3}{4}\), \(b = -2\), and \(c = \frac{1}{2}\).

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Most popular questions from this chapter

Find all real or imaginary solutions to each equation. Use the method of your choice. $$q^{2}+6 q-7=0$$

Find the solutions to \(6 x^{2}+5 x-4=0 .\) Is the sum of your solutions equal to \(-\frac{b}{a}\) ? Explain why the sum of the solutions to any quadratic equation is \(-\frac{b}{a}\) (Hint: Use the quadratic formula.)

Find the exact solution \((s)\) to each problem. If the solution(s) are irrational, then also find approximate solution(s) to the nearest tenth. Tossing a ball. A ball is tossed into the air at \(10 \mathrm{ft} / \mathrm{sec}\) from a height of 5 feet. How long does it take to reach the earth?

Find the exact solution \((s)\) to each problem. If the solution(s) are irrational, then also find approximate solution(s) to the nearest tenth. Missing numbers. Find two positive real numbers that differ by 1 and have a product of 16.

Find the exact solution \((s)\) to each problem. If the solution(s) are irrational, then also find approximate solution(s) to the nearest tenth. Diagonal of a square. The diagonal of a square is \(2 \mathrm{m}\) longer than a side. Find the length of a side.
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