Problem 64
Question
Choose a method and solve the quadratic equation. Explain your choice. $$ x^{2}-2 x=2 $$
Step-by-Step Solution
Verified Answer
The solutions to the given quadratic equation are \(x = 1+sqrt(3)\) and \(x = 1-sqrt(3)\)
1Step 1: Understand the Given Equation
The given equation is \(x^{2}-2x-2 = 0\). This is a quadratic equation which has the form \(ax^2 + bx + c = 0\). Here, a = 1, b = -2, and c = -2.
2Step 2: Apply the Quadratic Formula
Use the quadratic formula \(x = [-b ± sqrt(b^2-4ac)] / 2a\). Substitute the coefficients a, b, and c into the quadratic formula. This gives \(x = [2 ± sqrt((-2)^2 - 4*1*(-2))] / 2*1 = [2 ± sqrt(4 + 8)] / 2 = [2 ± sqrt(12)] / 2 = [2 ± 2sqrt(3)] / 2\).
3Step 3: Simplify the solutions
When we simplify this, we get two solutions: \(x = 1+sqrt(3)\) and \(x = 1-sqrt(3)\)
Key Concepts
Quadratic FormulaSolutions of Quadratic EquationsSimplification of Solutions
Quadratic Formula
The quadratic formula is a critical tool in solving quadratic equations of the form \( ax^2 + bx + c = 0 \). This formula allows us to find the solutions, also known as the roots, of the quadratic equation. The formula is given by:
\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here’s how it works:
\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here’s how it works:
- The term \( -b \pm \sqrt{b^2 - 4ac} \) represents the calculation that leads to the potential solutions of the equation.
- The portion \( b^2 - 4ac \) is called the discriminant and determines the nature of the roots.
- Using \( 2a \) in the denominator dictates how these solutions are balanced through dividing, adding or subtracting the other values.
Solutions of Quadratic Equations
Quadratic equations typically have two solutions, which can often be real or complex numbers, depending on the discriminant \( b^2 - 4ac \). Here’s how to interpret these solutions:
- If \( b^2 - 4ac > 0 \), the equation has two distinct real solutions.
- If \( b^2 - 4ac = 0 \), both solutions are real and equal, leading to one repeated solution.
- If \( b^2 - 4ac < 0 \), the solutions are complex or imaginary numbers, involving the square root of a negative number.
- \( x = 1 + \sqrt{3} \)
- \( x = 1 - \sqrt{3} \)
Simplification of Solutions
Simplifying the results from the quadratic formula is the final step to reach the solutions. It involves careful arithmetic manipulation:
With the solutions \( x = \frac{2 \pm 2\sqrt{3}}{2} \), we perform simplification:
With the solutions \( x = \frac{2 \pm 2\sqrt{3}}{2} \), we perform simplification:
- First, note the common factor in the expressions, in this case, 2 in the numerator and denominator.
- By factoring, divide both the numerator and denominator by 2, resulting in \( x = 1 \pm \sqrt{3} \).
- Always look for common factors that can cancel out terms in the expression.
- Recognize that algebraic manipulation should not alter the magnitudes and directions of calculations but simplify the representation of the roots.
Other exercises in this chapter
Problem 63
Simplify the radical expression. $$\sqrt{60}$$
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Solve the equation by factoring. $$ x^{2}-25=0 $$
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Solve the equation. Write the solutions as integers if possible. Otherwise, write them as radical expressions. $$ x^{2}-16=-7 $$
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