the equation x2 +px+q=0 has roots alpha and beta.show that alpha3+beta3=3pq-q3.suppose that one root is the square of the other,then by considering ( alpha - beta2)(beta-alpha2)=0,show that q2 +q +p3-3pq=0.

Open in App

Solution

$Given, the equation x^{2} + px + q has roots α and β . So, α + β = - (\frac{coefficient of x}{coefficient of x^{2}}) = - \frac{p}{1} = - p αβ = (\frac{constant term}{coefficient of x^{2}}) = \frac{q}{1} = q Thus, α^{3} + β^{3} = {(α + β)}^{3} - 3 αβ (α + β) = - p^{3} - 3 q (- p) = 3 pq - p^{3} \Rightarrow α^{3} + β^{3} = 3 pq - p^{3} And, (α - β^{2}) (β - α^{2}) = αβ - β^{3} - α^{3} + α^{2} β^{2} = αβ - (β^{3} + α^{3}) + {(αβ)}^{2} = q - 3 pq + p^{3} + q^{2} = q + q^{2} + p^{3} - 3 pq Hence Proved .$

Suggest Corrections

Similar questions

α, a n d β

x^{2} + p x + q = 0

α^{3}, β^{3}

If one root is square of the other root of the equation $x^{2} + p x + q = 0$ , then the relation between p and q is

x^{2} + p x + q = 0,

The equation $x^{2} + p x - q = 0$ has one root as a square root of the other.
If $p^{3} + q^{2} = q (1 + k p)$ , find the value of k.

x^{2} + p x + q = 0

p^{3} + q^{2} + q =

Join BYJU'S Learning Program