(1)given that -2x^2+6x+7=0, if alfa and B are the roots of the equation, form new equation whose roots are(a)alfa/B and B/alfa (b)alfa/B^2and B/alfa^2(b) if alfa and B are the roots of the equation 4x^+8x-1=0. find the values of alfa^3 B+alfaB^3

Answer 1

x^2 - 3 x - 7/2 = 0

x = [ 3 +/-sqrt (9+14) ]/2

A = [ 3 + 23]/2 = 13
B = [ 3 - 23] / 2 = -10

(a)
(x- A/B)(x-B/A) = 0
(x +13/10)(x+10/13) = 0
x^2 + (269/130) x + 1 = 0
130 x^2 + 269 x + 130 = 0
etc etc etc

Answer 2

Or

using a property of quadratic roots:
for ax^2 + bx + c = 0
if p and q are the roots
p+q = -b/a
qp = c/a

so for the given equation:
p+q = -6/-2 = 3
pq = -7/2

a)
for new equation:
roots are p/q and q/p
sum = p/q + q/p
= (p^2+q^2)/(pq)
= ( (p+q)^2 - 2pq)/(pq)
= (9 - 2(-7/2))/(-7/2)
= 16(-2/7)
= -32/7
product = (p/q)(q/p) = 1 = 7/7

so the equation is 7x^2 + 32x + 7 = 0

( Damon's roots should have been:
A = [ 3 + √23]/2
B = [ 3 - √23] / 2 )

b) root of new equation are p/q^2 and q/p^2
new sum = p/q^2 + q/p^2
= (p^3 + q^3)/(pq)^2

tricky part here ...

(p + q)^3 = p^3 + 3p^2q + 3pq^2 + q^3
= p^3 + q^3 + 3pq(p+q)
3^3 = p^3 + q^3 + 3(-7/2)(3)
27 + 63/2 = p^3 + q^3
p^3 + q^3 = 117/2
new sum = (117/2) / (49/4)
= (117/2)(4/49)
= 234/49

new product = (p/q^2)(q/p^2)
= pq/(pq)2
= 1/pq
= -2/7

new equation:
x^2 - (234/49)x - 2/7 = 0
times 49
49x^2 - 234x - 14 = 0

Answer 3

. Using x^2 +(p+q)x +pq new equatin is x^2 -3p -7/2 =2x^2 -6x - 7. . But ur ques is nt clear