Problem 66
Question
Solve. Write answers in standard form. $$ 3 x-x^{2}=5 $$
Step-by-Step Solution
Verified Answer
The solutions, in standard form, are complex: \( x = \frac{3}{2} \pm \frac{i \sqrt{11}}{2} \).
1Step 1: Arrange Equation
First, rewrite the given equation in the standard form of a quadratic equation, which is \( ax^2 + bx + c = 0 \). Begin by moving all terms to one side of the equation.
2Step 2: Rearrange Terms
The equation \( 3x - x^2 = 5 \) becomes \( -x^2 + 3x - 5 = 0 \) after subtracting 5 from both sides.
3Step 3: Standard Form
Multiply the entire equation by -1 to express it in the standard quadratic form, especially emphasizing a positive leading coefficient: \( x^2 - 3x + 5 = 0 \).
4Step 4: Calculate the Discriminant
Use the discriminant formula \( b^2 - 4ac \), where \( a = 1 \), \( b = -3 \), and \( c = 5 \). Calculate the discriminant: \((-3)^2 - 4 \times 1 \times 5 = 9 - 20 = -11\).
5Step 5: Solve Using the Quadratic Formula
Since the discriminant is negative, use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Substituting the values gives \( x = \frac{3 \pm \sqrt{-11}}{2} \).
6Step 6: Simplify the Roots
Simplify the expression to \( x = \frac{3}{2} \pm \frac{i \sqrt{11}}{2} \). The solutions are complex numbers because of the negative discriminant.
Key Concepts
DiscriminantQuadratic FormulaComplex Numbers
Discriminant
When solving quadratic equations, one key element is the discriminant. This small but mighty part of the equation denotes whether the solutions of the equation are real or complex numbers.
For any quadratic equation in the form \( ax^2 + bx + c = 0 \), the discriminant is calculated as \( b^2 - 4ac \).
Here is how it affects your solutions:
For any quadratic equation in the form \( ax^2 + bx + c = 0 \), the discriminant is calculated as \( b^2 - 4ac \).
Here is how it affects your solutions:
- If the discriminant is positive, the quadratic equation has two distinct real roots.
- If the discriminant is zero, there is exactly one real root, and it's known as a repeated or double root.
- If the discriminant is negative, as in our example where \( -11 \) was the outcome, the roots are complex numbers. This means they will involve the imaginary unit \( i \), which stands for the square root of -1.
Quadratic Formula
The quadratic formula is a magic wand for solving any quadratic equation. This formula can bail you out even when factoring is tricky or impossible. It’s the go-to method especially when dealing with complex numbers.
The formula is given by:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]To use it, simply plug in the values for \( a \), \( b \), and \( c \) from your standard form quadratic equation \( ax^2 + bx + c = 0 \).
The plus-minus symbol (\( \pm \)) tells you that there will be two solutions or roots for all quadratic equations.
In our example, substituting into the quadratic formula, we get:\[x = \frac{3 \pm \sqrt{-11}}{2}\]The negative discriminant results in a square root of a negative number, hinting that the roots are not real but complex.
The formula is given by:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]To use it, simply plug in the values for \( a \), \( b \), and \( c \) from your standard form quadratic equation \( ax^2 + bx + c = 0 \).
The plus-minus symbol (\( \pm \)) tells you that there will be two solutions or roots for all quadratic equations.
In our example, substituting into the quadratic formula, we get:\[x = \frac{3 \pm \sqrt{-11}}{2}\]The negative discriminant results in a square root of a negative number, hinting that the roots are not real but complex.
Complex Numbers
Complex numbers extend the one-dimensional number line to the two-dimensional complex plane, allowing us to include all numbers. They are written in the form \( a + bi \), where \( a \) is the real part and \( b \) is the imaginary part.
The introduction of \( i \), the imaginary unit defined as \( \sqrt{-1} \), allows us to express roots of negative numbers.
In our quadratic equation, the solutions involved complex numbers because of a negative discriminant:
The introduction of \( i \), the imaginary unit defined as \( \sqrt{-1} \), allows us to express roots of negative numbers.
In our quadratic equation, the solutions involved complex numbers because of a negative discriminant:
- The computed solution came out as \( x = \frac{3}{2} \pm \frac{i \sqrt{11}}{2} \).
- This showed two complex solutions: \( x = \frac{3}{2} + \frac{i \sqrt{11}}{2} \) and \( x = \frac{3}{2} - \frac{i \sqrt{11}}{2} \).
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