Westonci.ca connects you with experts who provide insightful answers to your questions. Join us today and start learning! Join our platform to connect with experts ready to provide accurate answers to your questions in various fields. Join our Q&A platform to connect with experts dedicated to providing accurate answers to your questions in various fields.
Sagot :
To solve the problem, we start with the equation:
[tex]\[ x + \frac{1}{x} = 3 \][/tex]
We need to find the value of [tex]\( \frac{6x}{x^2 + 1} \)[/tex].
First, solve [tex]\( x + \frac{1}{x} = 3 \)[/tex]. Rewrite the equation to form a quadratic equation:
[tex]\[ x + \frac{1}{x} = 3 \implies x^2 + 1 = 3x \implies x^2 - 3x + 1 = 0 \][/tex]
We solve the quadratic equation [tex]\( x^2 - 3x + 1 = 0 \)[/tex] using the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
where [tex]\(a = 1\)[/tex], [tex]\(b = -3\)[/tex], and [tex]\(c = 1\)[/tex].
[tex]\[ x = \frac{3 \pm \sqrt{9 - 4}}{2} = \frac{3 \pm \sqrt{5}}{2} \][/tex]
Thus, the two solutions for [tex]\( x \)[/tex] are:
[tex]\[ x_1 = \frac{3 - \sqrt{5}}{2} \][/tex]
[tex]\[ x_2 = \frac{3 + \sqrt{5}}{2} \][/tex]
Next, we need to determine the value of [tex]\( \frac{6x}{x^2 + 1} \)[/tex] for each solution.
For [tex]\( x_1 = \frac{3 - \sqrt{5}}{2} \)[/tex]:
[tex]\[ x_1^2 = \left(\frac{3 - \sqrt{5}}{2}\right)^2 = \frac{(3 - \sqrt{5})^2}{4} = \frac{9 - 6\sqrt{5} + 5}{4} = \frac{14 - 6\sqrt{5}}{4} = \frac{7 - 3\sqrt{5}}{2} \][/tex]
[tex]\[ x_1^2 + 1 = \frac{7 - 3\sqrt{5}}{2} + 1 = \frac{7 - 3\sqrt{5} + 2}{2} = \frac{9 - 3\sqrt{5}}{2} \][/tex]
[tex]\[ \frac{6x_1}{x_1^2 + 1} = \frac{6 \left(\frac{3 - \sqrt{5}}{2}\right)}{\frac{9 - 3\sqrt{5}}{2}} = \frac{6 (3 - \sqrt{5})}{9 - 3\sqrt{5}} = \frac{18 - 6\sqrt{5}}{9 - 3\sqrt{5}} = \frac{6 (3 - \sqrt{5})}{3 (3 - \sqrt{5})} = 2 \][/tex]
Thus:
[tex]\[ \frac{6x_1}{x_1^2 + 1} = 2 \][/tex]
For [tex]\( x_2 = \frac{3 + \sqrt{5}}{2} \)[/tex]:
[tex]\[ x_2^2 = \left(\frac{3 + \sqrt{5}}{2}\right)^2 = \frac{(3 + \sqrt{5})^2}{4} = \frac{9 + 6\sqrt{5} + 5}{4} = \frac{14 + 6\sqrt{5}}{4} = \frac{7 + 3\sqrt{5}}{2} \][/tex]
[tex]\[ x_2^2 + 1 = \frac{7 + 3\sqrt{5}}{2} + 1 = \frac{7 + 3\sqrt{5} + 2}{2} = \frac{9 + 3\sqrt{5}}{2} \][/tex]
[tex]\[ \frac{6x_2}{x_2^2 + 1} = \frac{6 \left(\frac{3 + \sqrt{5}}{2}\right)}{\frac{9 + 3\sqrt{5}}{2}} = \frac{6 (3 + \sqrt{5})}{9 + 3\sqrt{5}} = \frac{18 + 6\sqrt{5}}{9 + 3\sqrt{5}} = \frac{6 (3 + \sqrt{5})}{3 (3 + \sqrt{5})} = 2 \][/tex]
Thus:
[tex]\[ \frac{6x_2}{x_2^2 + 1} = 2 \][/tex]
Although through careful simplification, it was found that both results amount to 2, it turns out the correct calculated was:
[tex]\[ \frac{6x_2}{x_2^2 + 1} = \frac{3\sqrt{5} + 9}{1 + ( \sqrt{5}/2 + 3/2) ^2} \][/tex]
[tex]\[ x + \frac{1}{x} = 3 \][/tex]
We need to find the value of [tex]\( \frac{6x}{x^2 + 1} \)[/tex].
First, solve [tex]\( x + \frac{1}{x} = 3 \)[/tex]. Rewrite the equation to form a quadratic equation:
[tex]\[ x + \frac{1}{x} = 3 \implies x^2 + 1 = 3x \implies x^2 - 3x + 1 = 0 \][/tex]
We solve the quadratic equation [tex]\( x^2 - 3x + 1 = 0 \)[/tex] using the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
where [tex]\(a = 1\)[/tex], [tex]\(b = -3\)[/tex], and [tex]\(c = 1\)[/tex].
[tex]\[ x = \frac{3 \pm \sqrt{9 - 4}}{2} = \frac{3 \pm \sqrt{5}}{2} \][/tex]
Thus, the two solutions for [tex]\( x \)[/tex] are:
[tex]\[ x_1 = \frac{3 - \sqrt{5}}{2} \][/tex]
[tex]\[ x_2 = \frac{3 + \sqrt{5}}{2} \][/tex]
Next, we need to determine the value of [tex]\( \frac{6x}{x^2 + 1} \)[/tex] for each solution.
For [tex]\( x_1 = \frac{3 - \sqrt{5}}{2} \)[/tex]:
[tex]\[ x_1^2 = \left(\frac{3 - \sqrt{5}}{2}\right)^2 = \frac{(3 - \sqrt{5})^2}{4} = \frac{9 - 6\sqrt{5} + 5}{4} = \frac{14 - 6\sqrt{5}}{4} = \frac{7 - 3\sqrt{5}}{2} \][/tex]
[tex]\[ x_1^2 + 1 = \frac{7 - 3\sqrt{5}}{2} + 1 = \frac{7 - 3\sqrt{5} + 2}{2} = \frac{9 - 3\sqrt{5}}{2} \][/tex]
[tex]\[ \frac{6x_1}{x_1^2 + 1} = \frac{6 \left(\frac{3 - \sqrt{5}}{2}\right)}{\frac{9 - 3\sqrt{5}}{2}} = \frac{6 (3 - \sqrt{5})}{9 - 3\sqrt{5}} = \frac{18 - 6\sqrt{5}}{9 - 3\sqrt{5}} = \frac{6 (3 - \sqrt{5})}{3 (3 - \sqrt{5})} = 2 \][/tex]
Thus:
[tex]\[ \frac{6x_1}{x_1^2 + 1} = 2 \][/tex]
For [tex]\( x_2 = \frac{3 + \sqrt{5}}{2} \)[/tex]:
[tex]\[ x_2^2 = \left(\frac{3 + \sqrt{5}}{2}\right)^2 = \frac{(3 + \sqrt{5})^2}{4} = \frac{9 + 6\sqrt{5} + 5}{4} = \frac{14 + 6\sqrt{5}}{4} = \frac{7 + 3\sqrt{5}}{2} \][/tex]
[tex]\[ x_2^2 + 1 = \frac{7 + 3\sqrt{5}}{2} + 1 = \frac{7 + 3\sqrt{5} + 2}{2} = \frac{9 + 3\sqrt{5}}{2} \][/tex]
[tex]\[ \frac{6x_2}{x_2^2 + 1} = \frac{6 \left(\frac{3 + \sqrt{5}}{2}\right)}{\frac{9 + 3\sqrt{5}}{2}} = \frac{6 (3 + \sqrt{5})}{9 + 3\sqrt{5}} = \frac{18 + 6\sqrt{5}}{9 + 3\sqrt{5}} = \frac{6 (3 + \sqrt{5})}{3 (3 + \sqrt{5})} = 2 \][/tex]
Thus:
[tex]\[ \frac{6x_2}{x_2^2 + 1} = 2 \][/tex]
Although through careful simplification, it was found that both results amount to 2, it turns out the correct calculated was:
[tex]\[ \frac{6x_2}{x_2^2 + 1} = \frac{3\sqrt{5} + 9}{1 + ( \sqrt{5}/2 + 3/2) ^2} \][/tex]
Thank you for your visit. We're dedicated to helping you find the information you need, whenever you need it. Thank you for your visit. We're dedicated to helping you find the information you need, whenever you need it. Thank you for using Westonci.ca. Come back for more in-depth answers to all your queries.
