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Chapter 7: Problem 61

Solve. $$2 x^{2}=5 x$$

Short Answer

Expert verified
\(x = 0\) or \(x = \frac{5}{2}\)

Step by step solution

01

Rewrite the Equation

Start by rewriting the given equation to set it equal to zero. This will turn it into a standard quadratic equation format.Given equation: \[2x^2 = 5x\]Subtract \(5x\) from both sides:\[2x^2 - 5x = 0\]
02

Factor Out the Common Term

Next, factor out the common term, which is \(x\), from the equation.\[2x^2 - 5x = x(2x - 5) = 0\]
03

Solve for Each Factor

Set each factor in the equation equal to zero and solve for \(x\).First factor:\[x = 0\]Second factor:\[2x - 5 = 0\]Solve for \(x\) in the second factor:\[2x = 5\]\[x = \frac{5}{2}\]

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Key Concepts

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

Factorization
Factorization is one of the primary methods for solving quadratic equations. It's a convenient way to simplify and solve equations when they can be expressed as a product of simpler polynomials. In our equation \2x^2 - 5x = 0\, we start by rewriting it in the standard quadratic form. Here, we take out the common term, x, which helps us to factorize:
  • \(2x^2 - 5x\) transforms into \(x(2x - 5) = 0\).
  • This breaks the equation into two simpler equations: \(x = 0\) and \(2x - 5 = 0\).
Factorization works best on quadratic equations that can easily be split into two binomial expressions. If you can spot common terms or use techniques like grouping, you'll find solving the equation a lot easier.
Quadratic Formula
When an equation cannot be easily factorized, the quadratic formula is an excellent tool. The quadratic formula is:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]For any quadratic equation in the form \(ax^2 + bx + c = 0\), this formula will give you the roots directly.
  • In our original problem, \(2x^2 - 5x = 0\), if you couldn't factorize it, you would set \(a = 2\), \(b = -5\), and \(c = 0\).
  • Substituting these values in the formula, you would get:\[x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(2)(0)}}{2(2)} = \frac{5 \pm \sqrt{25}}{4}\]
  • This simplifies to \(x = 0\) or \(x = \frac{5}{2}\), matching the solutions we found via factorization.
The quadratic formula is versatile and works for all quadratic equations, including those that do not factorize easily.
Roots of Equations
The solutions to a quadratic equation are also known as the roots of the equation. These roots are the values of \(x\) that make the equation true. In our case \2x^2 - 5x = 0\, the roots are the values of \(x\) where the equation equals zero.
  • When we factorized, we found: \(x = 0\) and \(x = \frac{5}{2}\).
  • These values satisfy the original equation, meaning if you substitute them back into \2x^2 - 5x\, you get zero.
Understanding roots is fundamental in algebra because it helps in solving equations, analyzing graphs, and even in areas like calculus. Different quadratic equations can have:
  • Two distinct real roots
  • One real root (a double root)
  • Complex roots
In our example, we found two distinct real roots, clearly illustrating these concepts.

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