Chapter 5: Problem 6
Find the exact solutions by using the Quadratic Formula. \(x^{2}-2 x-2=0\)
Short Answer
Expert verified
The exact solutions are \( x = 1 + \sqrt{3} \) and \( x = 1 - \sqrt{3} \).
Step by step solution
01
Identify Coefficients
The quadratic equation is in the form \( ax^{2} + bx + c = 0 \). In this equation, \( a = 1 \), \( b = -2 \), and \( c = -2 \).
02
Write the Quadratic Formula
The Quadratic Formula is \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). This formula will be used to find the solutions of the given quadratic equation.
03
Calculate the Discriminant
Calculate the discriminant \(b^2 - 4ac\). Here, \( b^2 = (-2)^2 = 4 \) and \( 4ac = 4 \times 1 \times (-2) = -8 \). Thus, the discriminant is \( 4 + 8 = 12 \).
04
Substitute into the Quadratic Formula
Use the values of \( a \), \( b \), and the discriminant in the Quadratic Formula. The formula becomes: \[ x = \frac{-(-2) \pm \sqrt{12}}{2 \times 1} \].
05
Simplify the Solution
Simplify \( \frac{2 \pm \sqrt{12}}{2} \). \( \sqrt{12} = \sqrt{4 \times 3} = 2\sqrt{3} \). Thus, the equation becomes \[ x = \frac{2 \pm 2\sqrt{3}}{2} \].
06
Final Answer
Divide both terms inside the fraction by 2: \[ x = 1 \pm \sqrt{3} \]. So, the solutions are \( x = 1 + \sqrt{3} \) and \( x = 1 - \sqrt{3} \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Discriminant
The discriminant is a key component of the quadratic formula. It is found inside the square root portion of the formula and is denoted as \( b^2 - 4ac \). The discriminant helps to determine the nature of the roots of a quadratic equation.
- If the discriminant is positive, as in our example with a value of 12, the quadratic equation will have two distinct real solutions.
- If the discriminant is zero, the equation has exactly one real solution, meaning both roots are the same, also known as a repeated root.
- If the discriminant is negative, the solutions are complex or imaginary numbers, as the square root of a negative number is not a real number.
Quadratic Equation
A quadratic equation is a polynomial equation of the second degree, which means it includes a variable raised to the power of two. The general form of a quadratic equation is \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants, and \( a eq 0 \).
These equations are called 'quadratic' because the highest power of the variable (\( x \)) is 2, which forms the shape of a parabolic curve when graphed.
To find the roots or solutions of a quadratic equation, we often use the Quadratic Formula, which is \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). This formula allows us to find the solutions for any quadratic equation by substituting the appropriate values of \( a \), \( b \), and \( c \). The quadratic formula is always applicable; however, remember that the discriminant's sign will affect the nature of the solutions.
These equations are called 'quadratic' because the highest power of the variable (\( x \)) is 2, which forms the shape of a parabolic curve when graphed.
To find the roots or solutions of a quadratic equation, we often use the Quadratic Formula, which is \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). This formula allows us to find the solutions for any quadratic equation by substituting the appropriate values of \( a \), \( b \), and \( c \). The quadratic formula is always applicable; however, remember that the discriminant's sign will affect the nature of the solutions.
Exact Solutions
When we solve a quadratic equation precisely without approximations, we refer to the results as exact solutions. The quadratic formula provides exact solutions in terms of radicals rather than decimal approximations.
- In our example, substituting into the quadratic formula and simplifying the expression \( \frac{2 \pm 2\sqrt{3}}{2} \) gives us the exact solutions \( x = 1 + \sqrt{3} \) and \( x = 1 - \sqrt{3} \).
- These solutions are exact because they preserve the radical form, \( \sqrt{3} \), instead of converting it into a decimal, which would result in an approximation.