If α, β are the roots of x^2 + ax + b = 0 and α + δ, β + δ are the roots of x^2 + px + q = 0 then show that a^2 − p^2 = 4(b − q).


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Question Number 202193 by MATHEMATICSAM last updated on 22/Dec/23

$If α, β are the roots of x^{2} + a x + b = 0$ $and α + δ, β + δ are the roots of$ $x^{2} + p x + q = 0 then show that$ $a^{2} - p^{2} = 4 (b - q) .$
	Answered by Rasheed.Sindhi last updated on 22/Dec/23
	${ ((x^2 + ax + b = 0 ; α+β =−a,αβ=b)),((x^2 + px + q = 0 ;^★ (α + δ)+(β + δ)=−p,^⧫ (α + δ)(β + δ)=q)) :} ^★ α+β+2δ=−p⇒−a+2δ=−p⇒a−p=2δ⇒δ=((a−p)/2) ^⧫ αβ+(α+β)δ+δ^2 =q⇒b−aδ+δ^2 ⇒b−q=aδ−δ^2 ⇒b−q=a(((a−p)/2))−(((a−p)/2))^2 ⇒b−q=(((a−p)/2))(a−((a−p)/2))=(((a−p)/2))(((a+p)/2)) ⇒4(b−q)=a^2 −p^2 QED$
	${\begin{cases} x^{2} + a x + b = 0; α + β = - a, α β = b \\ x^{2} + p x + q = 0;^{★} (α + δ) + (β + δ) = - p,^{⧫} (α + δ) (β + δ) = q \end{cases}$ $^{★} α + β + 2 δ = - p \Rightarrow - a + 2 δ = - p \Rightarrow a - p = 2 δ \Rightarrow δ = \frac{a - p}{2}$ $^{⧫} α β + (α + β) δ + δ^{2} = q \Rightarrow b - a δ + δ^{2} \Rightarrow b - q = a δ - δ^{2}$ $\Rightarrow b - q = a (\frac{a - p}{2}) - {(\frac{a - p}{2})}^{2}$ $\Rightarrow b - q = (\frac{a - p}{2}) (a - \frac{a - p}{2}) = (\frac{a - p}{2}) (\frac{a + p}{2})$ $\Rightarrow 4 (b - q) = a^{2} - p^{2}$ $QED$
	Answered by MM42 last updated on 22/Dec/23

	$(α + δ) + (β + δ) = - p \Rightarrow δ = \frac{a - p}{2}$ $(α + δ) (β + δ) = α β + (α + β) δ + δ^{2}$ $q = b - a δ + δ^{2} = b - a (\frac{a - p}{2}) + {(\frac{a - p}{2})}^{2}$ $4 q = 4 b - 2 a^{2} + 2 a p + a^{2} - 2 a p + p^{2}$ $\Rightarrow a^{2} - p^{2} = 4 (b - q) ✓$
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