Answer: q²x² - p³x + 3pqx + q = 0
Step-by-step explanation:
Since a and B are the roots of x² - px + q = 0, then x = a and x = B
⇒ (x - a)(x - B) = 0
⇒ x² - ax - Bx + aB = 0
⇒ x² - (a + B)x + aB = 0
Comparing this to the given equation of x² - px + q = 0, we discover that
Next, given the roots for the new equation as [tex]x = \dfrac{a}{B^2} \quad\text{and}\quad\ x=\dfrac{B}{a^2}[/tex]
⇒ [tex]\bigg(x - \dfrac{a}{B^2}\bigg)\bigg(x-\dfrac{B}{a^2}\bigg)=0[/tex]
⇒ (B²x - a)(a²x - B) = 0
⇒ a²B²x² - a³x - B³x + aB = 0
⇒ a²B²x² - (a³ + B³)x + aB = 0
Let's look at each term individually:
1st term: a²B²x² = (aB)²x² = q²x² (since q = aB)
2nd term: (a³ + B³)x cubic formula can be used
= [(a + B)(a² - aB + B²)]x
= [(a + B)(a² + (2aB - 3aB) + B²)]x
= [(a + B)(a² + 2aB + B² - 3aB)]x
= [(a + B)(a + B)² - 3aB)]x
= [( p )( p² - 3 q )]x since p = a + B and q = aB
= (p³ - 3pq)x distributed "p" into p² - 3q
= p³x - 3pqx distributed "x" into p³ - 3pq
3rd term: aB = q since q = aB
Put it all together:
q²x² - (p³x - 3pqx) + q = 0
⇒ q²x² - p³x + 3pqx + q = 0
