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Chapter 1: Problem 125

When the sum of 6 and twice a positive number is subtracted from the square of the number, 0 results. Find the number.

Short Answer

Expert verified
The solutions to the problem are \(x = 1 + sqrt{7}\) and \(x = 1 - sqrt{7}\), but since x has to be a positive number, the solution must be \(x = 1 + sqrt{7}\), which approximates to 3.65

Step by step solution

01

Translation to Equation

The problem can be translated into the following quadratic equation: \(x^2 - (6 + 2x) = 0\). Where x is the positive number that the problem is asking to find.
02

Simplifying Equation

After simplifying the above equation, it becomes: \(x^2 - 2x - 6 = 0\)
03

Solving the Quadratic Equation

By using the quadratic formula \(x = [-(-2)卤sqrt{(-2)^2 - (4)(1)(-6)}]/2(1)\). This simplifies further to \(x = (2卤sqrt{4 + 24})/2\), which gives us two possible solutions: \(x = (2卤sqrt{28})/2 = 1卤sqrt{7}\)

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Most popular questions from this chapter

Exercises \(177-179\) will help you prepare for the material covered in the next section. Factor completely: \(x^{3}+x^{2}-4 x-4\)

Explain how to solve \(x^{2}+6 x+8=0\) using the quadratic formula.

$$\text { Solve for } C: \quad V=C-\frac{C-S}{L} N$$

If a quadratic equation has imaginary solutions, how is this shown on the graph of \(y=a x^{2}+b x+c ?\)

Solve each formula for the specified variable. Do you recognize the formula? If so, what does it describe? $$V=\frac{1}{3} B h \text { for } B$$
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