Answer: (i) α² + β² = 3/5
[tex]\bold{(ii)\ \alpha^{-1}=\dfrac{-5-\sqrt5}{2}\qquad \beta^{-1}=\dfrac{-5+\sqrt5}{2}}[/tex]
Step-by-step explanation:
p(x) = 5x² + 5x + 1
Use the quadratic formula to find the zeros:
[tex]x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}\\\\\\x=\dfrac{-5\pm \sqrt{5^2-4(5)(1)}}{2(5)}\\\\\\.\quad =\dfrac{-5\pm \sqrt{25-20}}{10}\\\\\\.\quad =\dfrac{-5\pm \sqrt5}{10}\\\\\\\alpha=\dfrac{-5+ \sqrt5}{10}\qquad \qquad \qquad \qquad \beta=\dfrac{-5+ \sqrt5}{10}[/tex]
[tex]\alpha^2=\bigg(\dfrac{-5+ \sqrt5}{10}\bigg)^2\qquad \qquad \quad\beta^2=\bigg(\dfrac{-5- \sqrt5}{10}\bigg)^2\\\\\\\alpha^2=\dfrac{25-10\sqrt5 +5}{100}\qquad \qquad \beta^2=\dfrac{25+10\sqrt5 +5}{100}\\\\\\\alpha^2=\dfrac{30-10\sqrt5}{100}\qquad \qquad \beta^2=\dfrac{30+10\sqrt5}{100}\\\\\\\alpha^2=\dfrac{3-\sqrt5}{10}\qquad \qquad \beta^2=\dfrac{3+\sqrt5}{10}\\\\\\\alpha^2+\beta^2=\dfrac{3-\sqrt5}{10}+\dfrac{3+\sqrt5}{10}\quad =\dfrac{6}{10}\quad =\large\boxed{\dfrac{3}{5}}[/tex]
[tex]\alpha^{-1}=\dfrac{10}{-5+ \sqrt5}\bigg(\dfrac{-5-\sqrt5}{-5-\sqrt5}\bigg)=\dfrac{10(-5-\sqrt5)}{20}\quad =\large\boxed{\dfrac{-5-\sqrt5}{2}}\\ \\\\\\\beta^{-1}=\dfrac{10}{-5- \sqrt5}\bigg(\dfrac{-5+\sqrt5}{-5+\sqrt5}\bigg)=\dfrac{10(-5+\sqrt5)}{20}\quad =\large\boxed{\dfrac{-5+\sqrt5}{2}}[/tex]
