Answer:
7
Step-by-step explanation:
1. Solve for x in [tex]x^2 = 3x - 1[/tex].
[tex]x=\frac{3+ \sqrt{5}}{2} ,\frac{3- \sqrt{5}}{2}[/tex]
2. Substitute [tex]x=\frac{3+ \sqrt{5}}{2} ,\frac{3- \sqrt{5}}{2}[/tex] into [tex]x^2 + \frac{1}{x^2} = y[/tex].
[tex]\frac{(3+\sqrt{5})^2}{(2,2(3-\sqrt{5}))^2} + \frac{(2,2(3-\sqrt{5}))^2}{(3 + \sqrt{5})^2} = y[/tex]
3. Solve for y in [tex]\frac{(3+\sqrt{5})^2}{(2,2(3-\sqrt{5}))^2} + \frac{(2,2(3-\sqrt{5}))^2}{(3 + \sqrt{5})^2} = y[/tex]
[tex]y=\frac{(3+\sqrt{5})^2}{(2,2(3-\sqrt{5}))^2} + \frac{(2,2(3-\sqrt{5}))^2}{(3 + \sqrt{5})^2}[/tex]
4. Therefore,
[tex]x=\frac{3 + \sqrt{5} }{2}, \frac{3-\sqrt{5}}{2}\\y=\frac{(3+\sqrt{5})^2}{(2,2(3-\sqrt{5}))^2} + \frac{(2,2(3-\sqrt{5}))^2}{(3 + \sqrt{5})^2}[/tex]
............
[tex]x= \frac{3-\sqrt{5}}{2}[/tex]≈[tex]0.381966011, y=7[/tex]
[tex]x= \frac{\sqrt{5}+3}{2}[/tex]≈[tex]2.618033989, y=7[/tex]
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Answer:
7
Step-by-step explanation:
Given equation: [tex]x^2=3x-1[/tex]
Rearrange so that it is equal to zero:
[tex]\implies x^2-3x+1=0[/tex]
Quadratic formula
[tex]x=\dfrac{-b \pm \sqrt{b^2-4ac} }{2a}\quad\textsf{when}\:ax^2+bx+c=0[/tex]
Using the quadratic formula to solve the given equation:
[tex]\implies x=\dfrac{-(-3) \pm \sqrt{(-3)^2-4(1)(1)} }{2(1)}[/tex]
[tex]\implies x=\dfrac{3 \pm \sqrt{5}}{2}[/tex]
[tex]\implies x^2=\dfrac{7 \pm 3\sqrt{5}}{2}[/tex]
Substituting this into [tex]x^2+\dfrac{1}{x^2}[/tex]
[tex]\implies \dfrac{7\pm3\sqrt{5}}{2}+\dfrac{1}{\dfrac{7\pm3\sqrt{5}}{2}}[/tex]
[tex]\implies \dfrac{7\pm3\sqrt{5}}{2}+\dfrac{2}{7\pm3\sqrt{5}}[/tex]
[tex]\implies \dfrac{(7\pm3\sqrt{5})(7\pm3\sqrt{5})}{2(7\pm3\sqrt{5})}+\dfrac{4}{2(7\pm3\sqrt{5})}[/tex]
[tex]\implies \dfrac{(7\pm3\sqrt{5})(7\pm3\sqrt{5})+4}{2(7\pm3\sqrt{5})}[/tex]
[tex]\implies \dfrac{49\pm42\sqrt{5}+45+4}{2(7\pm3\sqrt{5})}[/tex]
[tex]\implies \dfrac{98\pm42\sqrt{5}}{2(7\pm3\sqrt{5})}[/tex]
[tex]\implies \dfrac{14(7\pm3\sqrt{5})}{2(7\pm3\sqrt{5})}[/tex]
[tex]\implies \dfrac{14}{2}[/tex]
[tex]\implies 7[/tex]
