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Chapter 8: Problem 30

Solve. (Find all complex-number solutions.) \(5 x(x-1)-7=4 x(x-2)\)

Short Answer

Expert verified
The solutions are \( x = \frac{-3 + \sqrt{37}}{2} \) and \( x = \frac{-3 - \sqrt{37}}{2} \).

Step by step solution

01

Expand both sides of the equation

Expand the terms in parentheses: Left Side: \[5x(x - 1) - 7 = 5x^2 - 5x - 7\]Right Side: \[4x(x - 2) = 4x^2 - 8x\]So the equation becomes:\[5x^2 - 5x - 7 = 4x^2 - 8x\]
02

Simplify the equation

Subtract \(4x^2 - 8x\) from both sides: \[5x^2 - 5x - 7 - 4x^2 + 8x = 0\]Combine like terms: \[(5x^2 - 4x^2) + (-5x + 8x) - 7 = 0\]\[x^2 + 3x - 7 = 0\]
03

Apply the quadratic formula

To find the values of \(x\), use the quadratic formula: \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]For the equation \(x^2 + 3x - 7 = 0\), \(a = 1\), \(b = 3\), and \(c = -7\). Substitute these values into the formula: \[x = \frac{-3 \pm \sqrt{3^2 - 4(1)(-7)}}{2(1)}\]\[x = \frac{-3 \pm \sqrt{9 + 28}}{2}\]\[x = \frac{-3 \pm \sqrt{37}}{2}\]
04

Simplify the solutions

Thus, the solutions are: \[x = \frac{-3 + \sqrt{37}}{2}\] and \[x = \frac{-3 - \sqrt{37}}{2}\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Complex Number Solutions
When dealing with quadratic equations, sometimes the solutions can be complex numbers. A complex number is a combination of a real part and an imaginary part, represented as \( a + bi \), where \( i \) is the imaginary unit and equals the square root of -1. In this exercise, however, our solutions \( \frac{-3 + \sqrt{37}}{2} \) and \( \frac{-3 - \sqrt{37}}{2} \) are real numbers, because the discriminant \( b^2 - 4ac \) is positive. If the discriminant were negative, the solutions would involve imaginary numbers, making them complex.
Quadratic Formula
The quadratic formula is a powerful tool for solving quadratic equations of the form \( ax^2 + bx + c = 0 \). The formula is:
\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).
This formula allows us to find the solutions (roots) of any quadratic equation by simply plugging in the coefficients \(a\), \(b\), and \(c\). In our case, the quadratic equation was \( x^2 + 3x - 7 = 0 \) with \( a = 1 \), \( b = 3 \), and \( c = -7 \). By substituting these values into the quadratic formula, you can solve for \( x \) and find the roots of the equation.
Expanding Polynomials
Expanding polynomials involves multiplying each term inside a polynomial by every other term, which is an essential skill in algebra. In this exercise, we need to expand both sides of the equation.
For the left side (\( 5x(x-1) - 7 \)), we multiply 5x by both \( x \) and \( -1 \) to get \( 5x^2 - 5x \), and then subtract 7 to get \( 5x^2 - 5x - 7 \).
On the right side \( 4x(x-2) \), we multiply 4x by both \( x \) and \( -2 \) to get \( 4x^2 - 8x \). This step is crucial before you can combine like terms and simplify the equation.
Simplifying Equations
Simplifying equations is essential to find the solutions efficiently. Once the equation is expanded, we combine like terms and move them to one side.
From the expanded form \( 5x^2 - 5x - 7 = 4x^2 - 8x \), we subtract \( 4x^2 - 8x \) from both sides to get:
\( 5x^2 - 5x - 7 - 4x^2 + 8x = 0 \).
Combining like terms yields \( x^2 + 3x - 7 = 0 \). Now, this simplified form allows us to apply the quadratic formula easily. Simplifying equations makes solving them much more straightforward and manageable.

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Most popular questions from this chapter

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