Question

If α, β are zeroes of a polynomial p(x) = 5x² + 5x + 1, find

(i) α² + β²

(ii) α⁻¹ and β⁻¹

Answer 1

Answer: (i) α² + β² = 3/5

`\bold{(ii)\ \alpha^(-1)=(-5-\sqrt5)/(2)\qquad \beta^(-1)=(-5+\sqrt5)/(2)}`

Explanation:

p(x) = 5x² + 5x + 1

Use the quadratic formula to find the zeros:

`x=(-b\pm √(b^2-4ac))/(2a)\\\\\\x=(-5\pm √(5^2-4(5)(1)))/(2(5))\\\\\\.\quad =(-5\pm √(25-20))/(10)\\\\\\.\quad =(-5\pm \sqrt5)/(10)\\\\\\\alpha=(-5+ \sqrt5)/(10)\qquad \qquad \qquad \qquad \beta=(-5+ \sqrt5)/(10)`

`\alpha^2=\bigg((-5+ \sqrt5)/(10)\bigg)^2\qquad \qquad \quad\beta^2=\bigg((-5- \sqrt5)/(10)\bigg)^2\\\\\\\alpha^2=(25-10\sqrt5 +5)/(100)\qquad \qquad \beta^2=(25+10\sqrt5 +5)/(100)\\\\\\\alpha^2=(30-10\sqrt5)/(100)\qquad \qquad \beta^2=(30+10\sqrt5)/(100)\\\\\\\alpha^2=(3-\sqrt5)/(10)\qquad \qquad \beta^2=(3+\sqrt5)/(10)\\\\\\\alpha^2+\beta^2=(3-\sqrt5)/(10)+(3+\sqrt5)/(10)\quad =(6)/(10)\quad =\large\boxed{(3)/(5)}`

`\alpha^(-1)=(10)/(-5+ \sqrt5)\bigg((-5-\sqrt5)/(-5-\sqrt5)\bigg)=(10(-5-\sqrt5))/(20)\quad =\large\boxed{(-5-\sqrt5)/(2)}\\ \\\\\\\beta^(-1)=(10)/(-5- \sqrt5)\bigg((-5+\sqrt5)/(-5+\sqrt5)\bigg)=(10(-5+\sqrt5))/(20)\quad =\large\boxed{(-5+\sqrt5)/(2)}`