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Magazine Chapter 9: Problem 89
Solve using the quadratic formula. $$ (2 x+3)(2 x-3)-5(x 2+1)=-9 $$
Short Answer
Expert verified
The solutions are \(x = \sqrt{5}i\) and \(x = -\sqrt{5}i\).
Step by step solution
01
Expand Both Sides of the Equation
Begin by expanding the left-hand side of the equation. First, expand \((2x + 3)(2x - 3) = (2x)^2 - (3)^2 = 4x^2 - 9\).Next, expand \(-5(x^2 + 1) = -5x^2 - 5\).This makes the expression: \[(4x^2 - 9) - (5x^2 + 5) = -9\].
02
Simplify the Equation
Combine like terms in the equation.\[4x^2 - 5x^2 - 9 - 5 = -9\].This simplifies to:\[-x^2 - 14 = -9\].
03
Rearrange into Standard Quadratic Form
Move all terms to one side to form a standard quadratic equation \(ax^2 + bx + c = 0\):\[-x^2 - 14 + 9 = 0\].This simplifies to:\[-x^2 - 5 = 0\].Which can be rearranged to:\[x^2 + 5 = 0\].
04
Solve Using the Quadratic Formula
The quadratic formula is \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\).Identify the coefficients from the equation \(x^2 + 0x + 5 = 0\):\(a = 1, b = 0, c = 5\).Substitute these into the formula:\[x = \frac{-0 \pm \sqrt{0^2 - 4 \times 1 \times 5}}{2 \times 1} = \frac{\pm \sqrt{-20}}{2}\].
05
Simplify the Solution
Calculate \(\sqrt{-20}\) to express it in terms of \(i\) (where \(i = \sqrt{-1}\)):\(\sqrt{-20} = \sqrt{4 \times 5} \cdot \sqrt{-1} = 2\sqrt{5}i\).Substitute back into the formula:\[x = \frac{\pm 2\sqrt{5}i}{2} = \pm \sqrt{5}i\].Thus, the solutions are \(x = \sqrt{5}i\) and \(x = -\sqrt{5}i\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Quadratic Formula
The quadratic formula is a powerful tool used to find the roots of any quadratic equation of the form \( ax^2 + bx + c = 0 \). It's highlighted by its general formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]This formula is helpful because it provides a way to solve quadratic equations even when they cannot be easily factored. Each part of the formula helps us to determine the roots:
- \(b^2 - 4ac\) is known as the discriminant. It tells us about the nature of the roots.
- If the discriminant is positive, there are two real roots.
- If the discriminant is zero, there is one real root (a repeated root).
- If the discriminant is negative, the equation has two complex roots.
Complex Numbers
Complex numbers often arise in solutions to quadratic equations, particularly when the discriminant is negative. A complex number consists of two parts: a real part and an imaginary part, represented as \(a + bi\), where:
- \(a\) is the real part.
- \(b\) is the imaginary part.
- \(i\) is the imaginary unit, defined as \(i = \sqrt{-1}\).
Algebraic Expansion
Algebraic expansion is the process of multiplying out expressions such as products of polynomials. In this exercise, expansion involves distributing terms and simplifying the expression:
- For \((2x + 3)(2x - 3)\), the difference of squares formula is used: \((a + b)(a - b) = a^2 - b^2\).
- This simplifies the expression to \(4x^2 - 9\).
- Similarly, \(-5(x^2 + 1)\) expands to \(-5x^2 - 5\).
Solving Equations
Solving equations involves finding the value(s) of variables that make the equation true. For a quadratic equation, this often involves manipulating and simplifying the expression to use the quadratic formula or other methods like factoring or completing the square.The process involves:
- Expanding and simplifying terms to bring the equation into a standard form \( ax^2 + bx + c = 0 \).
- Moving all terms to one side to ensure the equation is set to zero.
- Applying an appropriate solution method, in this case, the quadratic formula, to find the roots.
