The-roots-of-the-equation-2x-2-px-q-0-are-2-and-2-Find-the-values-of-p-and-q-

Question Number 145443 by 7770 last updated on 04/Jul/21

T h e r o o t s o f t h e e q u a t i o n

2 x^{2} + p x + q = 0 a r e 2 α + β a n d

α + 2 β . F i n d t h e v a l u e s o f p a n d q

Answered by Olaf_Thorendsen last updated on 05/Jul/21

S = x_{1} + x_{2} = (2 α + β) + (α + 2 β) = 3 (α + β)

P = x_{1} x_{2} = (2 α + β) (α + 2 β) = 2 α^{2} + 5 α β + 2 β^{2}

S = - \frac{b}{a} = - \frac{p}{2} = 3 (α + β)

\Rightarrow p = - 6 (α + β)

P = \frac{c}{a} = \frac{q}{2} = 2 α^{2} + 5 α β + 2 β^{2}

\Rightarrow q = 4 α^{2} + 10 α β + 4 β^{2}

Answered by Rasheed.Sindhi last updated on 05/Jul/21

\underline{2 x^{2} + p x + q = 0; 2 α + β & α + 2 β a r e r o o t s .}

2 {(2 α + β)}^{2} + p (2 α + β) + q \dots . (i)

2 {(α + 2 β)}^{2} + p (α + 2 β) + q \dots . (i i)

(i) - (i i) :

2 {{(2 α + β)}^{2} - {(α + 2 β)}^{2}} + p {(2 α + β) - (α + 2 β)} = 0

- 6 α^{2} + 6 β^{2} - p (α - β) = 0

p = \frac{- 6 α^{2} + 6 β^{2}}{α - β)} = \frac{- 6 (α^{2} - β^{2})}{α - β} = - 6 (α + β)

p = - 6 (α + β)

(i) \Rightarrow q = - 2 {(2 α + β)}^{2} - p (2 α + β)

= - 2 {(2 α + β)}^{2} - {- 6 (α + β)} (2 α + β)

= - 8 α^{2} - 8 α β - 2 β^{2} + 6 (2 α^{2} + 3 α β + β^{2})

q = 4 α^{2} + 10 α β + 4 β^{2}