Chapter 10: Problem 15
Solve each radical equation. $$\sqrt{2 x-5}=\sqrt{x+4}$$
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Chapter 10: Problem 15
Solve each radical equation. $$\sqrt{2 x-5}=\sqrt{x+4}$$
These are the key concepts you need to understand to accurately answer the question.
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In Exercises \(75-92,\) rationalize each denominator. Simplify, if possible. $$\frac{2 \sqrt{x}+\sqrt{y}}{\sqrt{y}-2 \sqrt{x}}$$
In Exercises \(105-112,\) add or subtract as indicated. Begin by rationalizing denominators for all terms in which denominators contain radicals. $$\sqrt[3]{25}-\frac{15}{\sqrt[3]{5}}$$
Determine whether each relation is a function. (Section 8.1, Example 2) a. \(\\{(-1,1),(1,1),(-2,4),(2,4)\\}\) b. \([(1,-1),(1,1),(4,-2),(4,2)]\)
Use a graphing utility to solve each radical equation. Graph each side of the equation in the given viewing rectangle. The equation's solution set is given by the \(x\) -coordinate(s) of the point (s) of intersection. Check by substitution. $$\begin{aligned} &\sqrt{x}+4=2\\\ &[-2,18,1] \text { by }[0,10,1] \end{aligned}$$
Divide using synthetic division: $$\left(4 x^{4}-3 x^{3}+2 x^{2}-x-1\right) \div(x+3)$$ (Section \(5.6,\) Example 5 )
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