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Chapter 7: Problem 97

Solve: \(x(2 x+9)=5\)

Short Answer

Expert verified
The solutions to the equation are \(x = -\frac{1}{2}\) and \(x = 5\).

Step by step solution

01

Expand the Left Side

First, you need to distribute \(x\) on the left side of the equation to get rid of the parenthesis. This leads to \(2x^2 + 9x = 5\).
02

Bring All Terms to One Side

Move all terms to the left side of the equation. You do this by subtracting 5 from both sides, leading to \(2x^2 + 9x - 5 = 0\).
03

Use Quadratic Equation

Now solve for \(x\) using the quadratic formula: \[x = -\frac{b \pm \sqrt{b^2 - 4ac}}{2a}\] In this case, \(a=2\), \(b=9\), and \(c=-5\). Substituting the given values and simplifying will generate the solutions for \(x\).

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Key Concepts

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

Distributing in Algebra
Distributing in algebra is a fundamental concept that involves eliminating parentheses by multiplying a single term across terms inside the parentheses. This step is often necessary when simplifying expressions or equations.
In the exercise, we are given the expression \[x(2x+9)\]which means we need to distribute \(x\) over \(2x+9\).
To distribute correctly:
  • Multiply \(x\) by the first term inside the parentheses: \(x \times 2x = 2x^2\).
  • Then, multiply \(x\) by the second term: \(x \times 9 = 9x\).
After distribution, the equation becomes \[2x^2 + 9x = 5\].
This step makes the equation easier to manage and sets the stage for the next phase of solving it.
Expanding Equations
Expanding equations means expressing the equation in a format where all algebraic terms are spread out and simplified. This process is vital in breaking down the equation into a solvable form, especially when dealing with quadratic equations.
Once you've distributed, the equation \[x(2x+9)=5\]becomes\[2x^2 + 9x = 5\].
Next, move all terms to one side of the equation to form a standard quadratic equation by subtracting 5 from both sides:\[2x^2 + 9x - 5 = 0\].
  • The goal is to rewrite it in the standard quadratic form:\[ax^2 + bx + c = 0\], where it clearly shows the coefficients \(a\), \(b\), and \(c\).
Ensuring a precise expanded form allows you to accurately use algebraic tools like the quadratic formula for solving.
Solving Quadratic Equations
Once an equation is in the quadratic form\[2x^2 + 9x - 5 = 0\],the next step is solving it for \(x\). Quadratic equations can be solved by several methods, but here, we use the quadratic formula: \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\].
For our equation, identify the coefficients:
  • \(a = 2\)
  • \(b = 9\)
  • \(c = -5\)
Plug these values into the quadratic formula:\[x = \frac{-9 \pm \sqrt{9^2 - 4 \times 2 \times (-5)}}{2 \times 2}\].
Simplify inside the square root and perform the arithmetic:
  • First computation inside the square root is:\(9^2 - 4 \times 2 \times (-5)\) which simplifies to \(81 + 40 = 121\).
  • Thus, \[x = \frac{-9 \pm \sqrt{121}}{4}=\frac{-9 \pm 11}{4}\].
Finally, solve for the two values of \(x\) by separating the plus and minus solutions:
  • \(x = \frac{-9 + 11}{4} = \frac{2}{4} = \frac{1}{2}\)
  • \(x = \frac{-9 - 11}{4} = \frac{-20}{4} = -5\)
These are the solutions for the quadratic equation.

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Most popular questions from this chapter

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