Chapter 2: Problem 70
Use the Quadratic Formula to solve the quadratic equation. \(x^{2}+6 x+10=0\)
Short Answer
Expert verified
The roots of the equation \(x^{2}+6 x+10=0\) are \(x= -3 + i\) and \(x = -3 -i\).
Step by step solution
01
Identify the coefficients
From the equation \(x^{2}+6 x+10=0\), we can see that the coefficients are \(a = 1\), \(b = 6\), and \(c = 10\).
02
Calculate the Discriminant
The discriminant is calculated using the formula \(D = b^2 - 4ac\). Substituting the values, we get \(D = (6)^2 - 4*1*10 = 36 - 40 = -4\).
03
Find the roots using the Quadratic formula
The roots of the equation can be found using the quadratic formula \(-b \pm \sqrt{b^2 - 4ac} \, / \, 2a\). Substituting the values of \(a\), \(b\), and \(D\) in the quadratic formula, we get \(x = [-6 \pm \sqrt{-4}] / (2*1)\). Since the discriminant is negative, the roots will be complex and can be written as \(x = -3 \pm i\sqrt{1}\). Hence, the roots of the equation are \(x= -3 + i\) and \(x = -3 -i\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Quadratic Equation
A quadratic equation is a second-degree polynomial equation in a single variable, which can be written in the standard form as \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants, and \( a eq 0 \). The variable \( x \) represents the unknown that we are trying to solve for.
These equations are called 'quadratic' because they're related to a square (quadratus in Latin). Quadratic equations often appear in algebra, and they can represent various real-world scenarios, such as projectile motion or optimization problems.
To solve a quadratic equation, one can use methods like factoring, completing the square, or the quadratic formula. The Quadratic Formula is a universally applicable method for finding the roots of any quadratic equation, written as:
These equations are called 'quadratic' because they're related to a square (quadratus in Latin). Quadratic equations often appear in algebra, and they can represent various real-world scenarios, such as projectile motion or optimization problems.
To solve a quadratic equation, one can use methods like factoring, completing the square, or the quadratic formula. The Quadratic Formula is a universally applicable method for finding the roots of any quadratic equation, written as:
- \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
Discriminant
The discriminant is a crucial part of solving quadratic equations, found in the quadratic formula under the square root sign: \( \sqrt{b^2 - 4ac} \). This expression, \( b^2 - 4ac \), is called the discriminant and is denoted by \( D \). The value of \( D \) helps determine the nature of the roots of a quadratic equation.
Depending on the value of the discriminant, you can predict:
Depending on the value of the discriminant, you can predict:
- **Positive Discriminant**: If \( D > 0 \), the quadratic equation has two distinct real roots.
- **Zero Discriminant**: If \( D = 0 \), the quadratic equation has exactly one real root, or in other words, two identical real roots (also called repeated roots).
- **Negative Discriminant**: If \( D < 0 \), the quadratic equation has two complex roots, which are complex conjugates of each other.
Complex Roots
When a quadratic equation has a negative discriminant, it means the equation has complex roots. Complex numbers are numbers that have a real part and an imaginary part. They are expressed in the form \( a + bi \), where \( i \) is the imaginary unit, defined by \( i^2 = -1 \).
In our example equation, \( x^2 + 6x + 10 = 0 \), we found the discriminant to be \( -4 \). As seen in the solution process, a negative discriminant results in an expression under the square root that involves an imaginary number. The equation's roots are calculated using:
In our example equation, \( x^2 + 6x + 10 = 0 \), we found the discriminant to be \( -4 \). As seen in the solution process, a negative discriminant results in an expression under the square root that involves an imaginary number. The equation's roots are calculated using:
- \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
- \( x = \frac{-6 \pm \sqrt{-4}}{2} = -3 \pm i \)