Quadratic Equation

Question

The quadratic equation 2x²-3x-4 = 0 has roots p and q.

i) Without solving the equation, determine: p³+q³

ii) Hence or otherwise find a quadratic equation with integer coefficients which has roots p³ and q³.

Answer 1

Very interesting problem. The quadratic formula says p = (-b + √(b² - 4ac))/2a and q = (-b - √(b² - 4ac))/2a so you can cube these to find p³ and q³

p³ turns out to be (-b³ + 3b²√(b² - 4ac) - 3b³ + 12abc + (b² - 4ac)^3/2)/8a³

and q³ = (-b³ - 3b²√(b² - 4ac) - 3b³ + 12abc - (b² - 4ac)^3/2)/8a³

So p³ + q³ = (-2b³ - 6b³ + 24abc)/8a³ = (-8b³ + 24abc)/8a³ = (-b³ + 3abc)/a³

In this case, a = 2, b = -3, c = -4 so p³ + q³ = (-(-3)³ + 3(2)(-3)(-4))/(2)³ = (27 + 72)/8 = 99/8

Regarding question 2, if the roots are p³ and q³ that means the factors are (x - p³)(x - q³) = 0. Multiplying that out gives us x² - (p³ + q³)x + p³q³ = 0 and that will result in integer coefficients as long as p and q are integers.

Answer 2

2p^2 - 3p - 4 = 0 and 2q^2 - 3q - 4 = 0 , since p and q are solutions.

subtracting,

2p^2 - 3p = 2q^2 - 3q

2p^2 - 2q^2 = 3p - 3q <--- adds 3p - 2q^2 to both sides

2(p+q)(p-q) = 3(p-q) <--- factors both sides completely

2(p+q) = 3 <--- divides both sides by p-1, assuming p and q are different

p+q = 3/2 <--- please label this equation ALPHA

p^3 + q^2 = (p+q)(p^2 - pq + q^2) by SOAP formula

= (3/2)(p^2 - pq + q^2) <--- substitutes equation ALPHA

= (3/2)( p^2 - p(3/2 - p) + (3/2-p)^2) <-- per equation ALPHA: q = 3/2 - p

= (3/2) ( p^2 - (3/2)p + p^2 + (9/4 - 3p + p^2)) : distributes p over (3/2-p) and FOILS

= (3/2) ( p^2 - (3/2)p + p^2 + 9/4 - 3p + p^2) :

= (3/2) (p^2 - (9/2)p + 2p^2 + 9/4) :

= (3/2) ( 3p^2 - (9/2)p + 9/4) :combines like terms

= (9/2)( p^2 - (3/2)p + 3/4) : factors out 3

= (9/4)(2p^2 - 3p + 3/2) : factors out 1/2

= (9/4)( 2p^2 - 3p - 4 + 11/2) : adds and subtracts 4

= (9/4)[( 2p^2 - 3p - 4) + 11/2 ] : associative property

= (9/4)[ 0 + 11/2 ] : p is a solution, so the quadratic term is zero

= 9/4 * 11/2

= 99/8