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Chapter 1: Problem 2

Solve each equation in Exercises \(1-14\) by factoring. $$x^{2}-13 x+36=0$$

Short Answer

Expert verified
The solutions are \(x=4\) and \(x=9\).

Step by step solution

01

Identifying the coefficients

Identify the values of \(a\), \(b\), and \(c\) in the equation \(ax^{2} + bx + c =0\). Here, \(a=1\), \(b=-13\), and \(c=36\).
02

Factoring the quadratic equation

Factor the quadratic equation. The factors of \(c\) such that their sum equals \(b\) are \( -4\) and \(-9\). When we consider these factors, the equation becomes \((x-4)(x-9)=0\).
03

Solving the factored equation

To solve for \(x\), set each factor equal to zero. This results in two equations, \(x-4=0\) and \(x-9=0\). Solving for \(x\) gives us the solutions \(x=4\) and \(x=9\).

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Most popular questions from this chapter

In Exercises \(69-72,\) use inspection to describe each inequality's solution set. Do not solve any of the inequalities. $$ (x-2)^{2}>0 $$

In Exercises \(21-28,\) divide and express the result in standard form. $$\frac{8 i}{4-3 i}$$

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. $$ \frac{-x-3}{x+2} \leq 0 $$

In Exercises \(29-44,\) perform the indicated operations and write the result in standard form. $$\sqrt{-12}(\sqrt{-4}-\sqrt{2})$$

Solve each quadratic inequality in Exercises \(1-28\) and graph the solution set on a real number line. Express each solution set in interval notation. $$ \left|x^{2}+6 x+1\right|>8 $$
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