Question

if alpha and beta are the zeroes of the quadratic polynomial f(x)=x'2-2x-+3,find a polynomial whose zeroes are alpha-1/alpha+1,beta-1/beta+1

Answer 1

Given equation :-f(x) = 2x² - 2x + 3a = 2, b = -2, c = 3A/Q,
It is given that α & β are the zeros of given polynomial equation, then find out
(α - 1)/(α + 1) = ?(β - 1)/(β + 1) = ?
We know that,
α + β =   -b/a = -(-2/2) = 1    ⇒ α + β =  1 ----------------------Eqn(i)
and, 
αβ = c/a = 3/2⇒ αβ = 3/2  -----------------------Eqn(ii)
We know that,
(α + β)² - (α - β)² = 4αβ⇒ 1 ² - (α - β)² = 4×3/2⇒ (α - β)² = 1 - 6 = -5α - β  = √-5 = i√5
 From Eqn(i)  and Eqn(ii)2α = 1 + i√5    
α = (1 + i√5)/2so,β = (1 - i√5)/2Now (α - 1)/(α + 1) [tex] \frac{\frac{1+i \sqrt{5} }{2}-1}{\frac{1+i \sqrt{5} }{2}+1} \\ = \frac{1+i \sqrt{5}-2 }{1+i \sqrt{5}+2} \\ = \frac{-1+i \sqrt{5} }{3+i \sqrt{5} } \\ = \frac{(-1+i \sqrt{5})(3-i \sqrt{5} ) }{(3+i \sqrt{5}) (3-i \sqrt{5} )} \\ = \frac{2+4i \sqrt{5} }{8} \\ = \frac{1+i2 \sqrt{5} }{4} [/tex]
Similarilly We can find (β-1)/(β+1)

Answer 2

(a) Magnesium chloride →MgCl2 (b) Calcium oxide →CaO (c) Copper nitrate →Cu(NO3)2 (d) Aluminium chloride →AlCl3 (e) Calcium carbonate →CaCO3

6.022×1023 102

× 0.051 molecules

= 3.011 × 1020 molecules of Al2O3 The number of aluminium ions (Al3+) present in one molecules of aluminium oxide is 2. Therefore, The number of aluminium ions (Al3+) present in 3.11 × 1020 molecules (0.051g) of aluminium oxide (Al2O3) = 2 × 3.011 × 1020 = 6.022 × 1020 1 mole of aluminium oxide (Al2O3) = 2 × 27 + 3 × 16 = 102g i.e., 102g of Al2O3 = 6.022 × 1023 molecules of Al2O3 Then, 0.051 g of Al2O3 contains =