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Chapter 1: Problem 11

Solve each equation in Exercises \(1-14\) by factoring. $$2 x(x-3)=5 x^{2}-7 x$$

Short Answer

Expert verified
The solutions of the equation are \(x = 0\) and \(x = 1/3\).

Step by step solution

01

Arrange the equation into standard form

Start by combining like terms on both sides to rearrange the equation into the standard form of a quadratic equation \(ax^2 + bx + c = 0\). Given equation is \(2x(x-3) = 5x^{2} - 7x\). After simplifying it we get \(2x^2 - 6x = 5x^2 - 7x\). Subtract \(2x^2\) and \(6x\) from both sides, which gives us the equation \(3x^2 - x = 0\) in standard form.
02

Factor the equation

The quadratic equation \(3x^2 - x = 0\) can be factored by taking the common factor \(x\) out which gives \(x(3x - 1) = 0\).
03

Solve for x

After we have factored the equation, set each factor equal to zero and solve for \(x\). This gives \(x = 0\) or \(3x - 1 = 0\) which simplifies further to \(x = 0\) or \(x = 1/3\).

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Most popular questions from this chapter

In Exercises \(9-20,\) find each product and write the result in standard form. $$-8 i(2 i-7)$$

Solve each rational inequality in Exercises \(29-48,\) and graph the solution set on a real number line. Express each solution set in interval notation. $$ \frac{x-4}{x+3}>0 $$

Solve each rational inequality in Exercises \(29-48,\) and graph the solution set on a real number line. Express each solution set in interval notation. $$ \frac{x}{x-3}>0 $$

Which one of the following is true? a. Some irrational numbers are not complex numbers. b. \((3+7 i)(3-7 i)\) is an imaginary number. c. \(\frac{7+3 i}{5+3 i}=\frac{7}{5}\) d. In the complex number system, \(x^{2}+y^{2}\) (the sum of two squares) can be factored as \((x+y i)(x-y i)\)

Write a rational inequality whose solution set is \((-\infty,-4)\) or \([3, \infty)\)
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