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Chapter 9: Problem 55

Solve each equation. See Sections 2.1 and 5.8 $$ x^{2}+6=5 x $$

Short Answer

Expert verified
The solutions are \(x = 2\) and \(x = 3\).

Step by step solution

01

Move All Terms to One Side

Start by rewriting the equation so that all terms are on one side. Subtract \(5x\) from both sides of the equation: \[ x^2 + 6 - 5x = 0 \] This gives us the equation \( x^2 - 5x + 6 = 0 \).
02

Factor the Quadratic Equation

We need to factor the quadratic equation \( x^2 - 5x + 6 = 0 \). To do this, find two numbers whose product is 6 and sum is -5. These numbers are -2 and -3. Therefore, the equation factors as: \[ (x - 2)(x - 3) = 0 \]
03

Solve for x

Set each factor equal to zero and solve for \(x\).1. \( x - 2 = 0 \) leads to \( x = 2 \).2. \( x - 3 = 0 \) leads to \( x = 3 \).Thus, the solutions are \( x = 2 \) and \( x = 3 \).

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Key Concepts

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

Factoring Quadratic Equations
Factoring is a technique used in algebra to break down expressions into simpler, more manageable parts. It鈥檚 especially useful when solving quadratic equations. Quadratic equations are in the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants. Here, our task is to find two numbers that multiply to give \(c\) (in our case, 6) while also adding to equal \(b\) (which is -5 in the exercise).
  • In "\(x^2 - 5x + 6\)", look for two numbers that multiply to 6, the constant term.
  • These numbers also need to add up to -5, the coefficient of \(x\).
  • The numbers -2 and -3 work because \((-2) \times (-3) = 6\) and \((-2) + (-3) = -5\).
Once you find these numbers, rewrite the quadratic as a product of two binomials: \((x - 2)(x - 3) = 0\). This step transforms a complex equation into something straightforward, making it easier to solve.
Solving Equations by Factoring
After factoring the quadratic equation, we are left with factors that can be set to zero. This is because, if a product is zero, at least one of the factors must be zero. This simple principle is known as the "zero product property."
  • Look at the factored form: \((x - 2)(x - 3) = 0\).
  • To solve for \(x\), set each binomial equal to zero.
For the factors:

1. First Factor

Set \(x - 2 = 0\). Solving this gives \(x = 2\).

2. Second Factor

Set \(x - 3 = 0\). From this, find \(x = 3\).
Therefore, the original quadratic equation \(x^2 - 5x + 6 = 0\) has solutions \(x = 2\) and \(x = 3\). Solving equations like this means locating the points where the quadratic graph intersects the \(x\)-axis.
Algebraic Manipulation
Algebraic manipulation involves rearranging equations and expressions using a variety of operations and properties to arrive at a solution. It is indispensable in solving equations, especially in reducing them to simpler forms suitable for factoring.
In our original equation, \(x^2 + 6 = 5x\), algebraic manipulation helped reposition terms:
  • Start by moving \(5x\) from the right-hand side to the left-hand side to gather all terms on one side.
  • This helps in viewing the equation as \(x^2 - 5x + 6 = 0\).
By placing all terms on one side, the equation takes a standard quadratic form. This setup makes it convenient to apply methods like factoring. Proper manipulation is important not just for quadratic equations but for any algebraic problem, as it sets the stage for solution techniques to be applied effectively.

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Most popular questions from this chapter

Solve. See Example 4. $$ \log _{9} x=\frac{1}{2} $$

Explain why the graph of the function \(y=\log _{b} x\) contains the point \((1,0)\) no matter what \(b\) is.

Add or subtract as indicated. See Section 6.2. $$ \frac{5}{y+1}-\frac{4}{y-1} $$

CONCEPT EXTENSIONS. Without using a calculator, explain which of \(\log 50^{-1}\) or \(\ln 50^{-1}\) must be larger and why.

If \(x=-2, y=0,\) and \(z=3,\) find the value of each expression. See Section 1.3 $$ \frac{x^{3}-2 y+z}{2 z} $$
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