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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

{\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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