Question

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Answer 1

Answer:

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Answer 2

Question:-

If α and β are the roots of the equation x² + px + q = 0 and α = β² , then prove that p³ + q² + q = 3pq

Answer:-

Given quadratic equation is x² + px + q = 0.

On comparing with standard form of a Quadratic equation , i.e., ax² + bx + c = 0,

Let ,

• a = 1
• b = p
• c = q

We know that,

Sum of the zeroes = - b/a

⟶ α + β = - p/1

⟶ α + β = - p

⟶ - α - β = p

Substitute the value of α from equation (1).

⟶ - α - β = p -- equation (1)

Product of the zeroes = c/a

⟶ αβ = q -- equation (2)

Also given that,

α = β² -- equation (3)

We have to prove :

p³ + q² + q = 3pq

Substitute the values of p and q from equations (1) & (2).

⟶ (- α - β)³ + (αβ)² + αβ = 3(- α - β)(αβ)

• (a - b)³ = a³ - b³ - 3ab(a - b)

⟶ (- α)³ - (β)³ - 3( - α)(β) (- α - β) + (αβ)² + αβ = 3 [ αβ ( - α) + αβ ( - β) ]

⟶ - α³ - β³ + 3αβ( - α - β) + (αβ)² + αβ = 3 [ - α²β - αβ² ]

⟶ - α³ - β³ - 3α²β - 3αβ² + (αβ)² + αβ = 3 ( - α²β - αβ² )

⟶ - (β²)³ - β³ - 3(β²)²(β) - 3(β²)(β²) + (β²)²(β)² + (β²)(β) = 3 [ - (β²)²(β) - (β²)(β²) ]

[ From equation (1) ]

⟶ - β⁶ - β³ - 3β⁵ - 3β⁴ + β⁶ + β³ = 3 ( - β⁵ - β⁴ )

⟶ - 3β⁵ - 3β⁴ = - 3β⁵ - 3β⁴

Hence Proved.