Chapter 1: Problem 68
Find all real solutions of the quadratic equation. $$5 x^{2}-7 x+5=0$$
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Chapter 1: Problem 68
Find all real solutions of the quadratic equation. $$5 x^{2}-7 x+5=0$$
These are the key concepts you need to understand to accurately answer the question.
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A square field in a certain state park is mowed around the edges every week. The rest of the field is kept unmowed to serve as a habitat for birds and small animals (see the figure). The field measures \(b\) feet by \(b\) feet, and the mowed strip is \(x\) feet wide. (a) Explain why the area of the mowed portion is \(b^{2}-(b-2 x)^{2}\) (b) Factor the expression in (a) to show that the area of the mowed portion is also \(4 x(b-x)\) CAN'T COPY THE GRAPH
(a) Find the radius of each circle in the pair, and the distance between their centers; then use this information to determine whether the circles intersect. (i) \((x-2)^{2}+(y-1)^{2}=9\) \((x-6)^{2}+(y-4)^{2}=16\) (ii) \(x^{2}+(y-2)^{2}=4\) \((x-5)^{2}+(y-14)^{2}=9\) (iii) \((x-3)^{2}+(y+1)^{2}=1\) \((x-2)^{2}+(y-2)^{2}=25\) (b) How can you tell, just by knowing the radii of two circles and the distance between their centers, whether the circles intersect? Write a short paragraph describing how you would decide this and draw graphs to illustrate your answer.
Factor the expression completely. $$x^{3}+3 x^{2}-x-3$$
The left-hand column in the table lists some common algebraic errors. In each case, give an example using numbers that show that the formula is not valid. An example of this type, which shows that a statement is false, is called a counterexample.$$\begin{array}{|c|c|} \hline \text { Algebraic error } & \text { Counterexample } \\ \hline \frac{1}{a}+\frac{1}{b}=\frac{1}{\sqrt{a}+b} & \frac{1}{2}+\frac{1}{2} \neq \frac{1}{2+2} \\ (a+b)^{2}=a^{2}+b^{2} & \\ \sqrt{a^{2}+b^{2}}=a+b & \\ \frac{a+b}{a}=b \\ \left(a^{3}+b^{3}\right)^{1 / 3}=a+b & \\ a^{m} / a^{n}=a^{m / n} & \\ a^{-1 / n}=\frac{1}{a^{n}} & \\ \hline \end{array}$$
State whether the given equation is true for all values of the variables. (Disregard any value that makes a denominator zero.) $$\frac{x+1}{y+1}=\frac{x}{y}$$
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