Problem 59
Question
\(x^{2}-3 x+3=0\)
Step-by-Step Solution
Verified Answer
The roots are \( x = \frac{3 + \sqrt{3}i}{2} \) and \( x = \frac{3 - \sqrt{3}i}{2} \).
1Step 1 - Identify the coefficients
In the quadratic equation, identify the values of the coefficients. The standard form of a quadratic equation is x^2 + bx + c = 0. For the given equation, we have: \(a = 1\), \(b = -3\), and \(c = 3\).
2Step 2 - Calculate the discriminant
The discriminant of a quadratic equation \(ax^2 + bx + c = 0\) is given by \(D = b^2 - 4ac\). Substitute the values of \(a\), \(b\), and \(c\) into this formula.\[ D = (-3)^2 - 4(1)(3) = 9 - 12 = -3\].
3Step 3 - Analyze the discriminant
Since the discriminant \(D\) is less than 0 (\(D = -3\)), the quadratic equation \(x^2 - 3x + 3 = 0\) has no real roots. Instead, it has two complex conjugate roots.
4Step 4 - Calculate the complex roots
Use the quadratic formula \(x = \frac{-b \pm \sqrt{D}}{2a}\) to find the complex roots. Here, substitute \(a = 1\), \(b = -3\), and \(D = -3\):\[x = \frac{-(-3) \pm \sqrt{-3}}{2(1)} = \frac{3 \pm \sqrt{3}i}{2}\]. Therefore, the roots are \(x = \frac{3 + \sqrt{3}i}{2}\) and \(x = \frac{3 - \sqrt{3}i}{2}\).
Key Concepts
Quadratic FormulaDiscriminantComplex RootsQuadratic Equation
Quadratic Formula
The quadratic formula is an essential tool for solving quadratic equations. A quadratic equation is any equation that can be written in the standard form:
\[ax^2 + bx + c = 0\]
Here, \(a\), \(b\), and \(c\) are constants, and \(x\) represents the variable we want to solve for. The quadratic formula is given by:
\[x = \frac{-b \pm \sqrt{D}}{2a}\]
where \(D = b^2 - 4ac\) is the discriminant of the quadratic equation. Using this formula, you can find the solutions for \(x\) by substituting the coefficients \(a\), \(b\), and \(c\) from your specific equation. If you follow the formula's steps accurately, you will solve any quadratic equation, whether it has real or complex roots.
\[ax^2 + bx + c = 0\]
Here, \(a\), \(b\), and \(c\) are constants, and \(x\) represents the variable we want to solve for. The quadratic formula is given by:
\[x = \frac{-b \pm \sqrt{D}}{2a}\]
where \(D = b^2 - 4ac\) is the discriminant of the quadratic equation. Using this formula, you can find the solutions for \(x\) by substituting the coefficients \(a\), \(b\), and \(c\) from your specific equation. If you follow the formula's steps accurately, you will solve any quadratic equation, whether it has real or complex roots.
Discriminant
The discriminant \(D\) of a quadratic equation helps determine the nature of its roots. It is calculated using the formula:
\[D = b^2 - 4ac\]
Here's what the value of the discriminant tells us:
\[D = (-3)^2 - 4(1)(3) = 9 - 12 = -3\]
Since \(D < 0\), this tells us that the quadratic equation \(x^2 - 3x + 3 = 0\) has two complex roots.
\[D = b^2 - 4ac\]
Here's what the value of the discriminant tells us:
- If \(D > 0\), the equation has two distinct real roots.
- If \(D = 0\), the equation has exactly one real root (a repeated root).
- If \(D < 0\), the equation has two complex roots.
\[D = (-3)^2 - 4(1)(3) = 9 - 12 = -3\]
Since \(D < 0\), this tells us that the quadratic equation \(x^2 - 3x + 3 = 0\) has two complex roots.
Complex Roots
When the discriminant \(D\) of a quadratic equation is negative, the equation has complex roots. Complex roots appear in conjugate pairs (same real part but opposite imaginary parts). The quadratic formula to find these roots is still the same:
\[x = \frac{-b \pm \sqrt{D}}{2a}\]
However, since \(D < 0\), we write \(D\) as a negative number inside the square root. This introduces an imaginary unit \(i\), where \(i = \sqrt{-1}\). For the given exercise:
\[D = -3\]
So, replacing \(D\) with \(-3\) and solving:
\[x = \frac{-(-3) \pm \sqrt{-3}}{2(1)} = \frac{3 \pm \sqrt{3}i}{2}\]
Thus, the roots are \(x = \frac{3 + \sqrt{3}i}{2}\) and \(x = \frac{3 - \sqrt{3}i}{2}\). These roots have the same real part (\(\frac{3}{2}\)) but different imaginary parts (\(\frac{\sqrt{3}}{2}i\) and \(-\frac{\sqrt{3}}{2}i\)).
\[x = \frac{-b \pm \sqrt{D}}{2a}\]
However, since \(D < 0\), we write \(D\) as a negative number inside the square root. This introduces an imaginary unit \(i\), where \(i = \sqrt{-1}\). For the given exercise:
\[D = -3\]
So, replacing \(D\) with \(-3\) and solving:
\[x = \frac{-(-3) \pm \sqrt{-3}}{2(1)} = \frac{3 \pm \sqrt{3}i}{2}\]
Thus, the roots are \(x = \frac{3 + \sqrt{3}i}{2}\) and \(x = \frac{3 - \sqrt{3}i}{2}\). These roots have the same real part (\(\frac{3}{2}\)) but different imaginary parts (\(\frac{\sqrt{3}}{2}i\) and \(-\frac{\sqrt{3}}{2}i\)).
Quadratic Equation
A quadratic equation is any polynomial equation of the second degree, meaning the highest power of the variable \(x\) in the equation is \(x^2\). The general form of a quadratic equation is:
\[ax^2 + bx + c = 0\]
Where \(a\), \(b\), and \(c\) are coefficients, and \(x\) is the unknown variable. Here are a few key points:
\[ax^2 + bx + c = 0\]
Where \(a\), \(b\), and \(c\) are coefficients, and \(x\) is the unknown variable. Here are a few key points:
- \(a\) should not be zero, as it would then no longer be a quadratic equation.
- The term \(ax^2\) ensures the 'quadratic' characteristic.
- The solutions to a quadratic equation can be found using factoring, completing the square, or the quadratic formula.
Other exercises in this chapter
Problem 58
\(p^{2}-14 p+49=0\)
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The NCAA rules for women's basketball state that the rim of the hoop shall be \(10 \mathrm{ft}\) above the floor. The threepoint line on the floor is \(19 \math
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The length of a rectangle is \(4 \mathrm{ft}\) less than twice its width. The area of the rectangle is \(576 \mathrm{ft}^{2}\). Find the length and width.
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\(w^{2}+4 w+3=0\)
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