If α and β are roots of the equation x2+p x+(3 p/4)=0, suchThat |α-β|=√10, then p belongs to the se :

Question

If α and β are roots of the equation x2+p x+(3 p/4)=0, suchThat |α-β|=√10, then p belongs to the se : (A) 2, - 5 (B) - 3, 2 (C) - 2, 5 (D) 3, - 5

Tardigrade Admin · Accepted Answer

Given α, β are roots of equation x2+p x+3 p / 4=0 (α+β)=-p ; α ⋅ β=3 p / 4 ⇒ (α-β)=√10 ⇒ (α-β)2=10 ⇒ (α+β)2-4 α β=10 p2-4(3 p / 4)=10 p2-3 p-10=0 p2-5 p+2 p-10=0 (p-5)(p+2)=0 p=-2,5 P ∈ -2,5

Q. If $α$ and $β$ are roots of the equation $x^{2} + p x + \frac{3 p}{4} = 0$ , suchThat $∣ α - β ∣ = 10$ , then $p$ belongs to the se :

{2, - 5}

{- 3, 2}

{- 2, 5}

{3, - 5}

Q. If α and β are roots of the equation x2+px+43p​=0, suchThat ∣α−β∣=10​, then p belongs to the se :

{2, - 5}

{- 3, 2}

{- 2, 5}

{3, - 5}

Given α,β are roots of equation x2+px+3p/4=0 (α+β)=−p;α⋅β=3p/4 ⇒(α−β)=10​ ⇒(α−β)2=10 ⇒(α+β)2−4αβ=10 p2−4(3p/4)=10 p2−3p−10=0 p2−5p+2p−10=0 (p−5)(p+2)=0 p=−2,5 P∈{−2,5}

Q. If $α$ and $β$ are roots of the equation $x^{2} + p x + \frac{3 p}{4} = 0$ , suchThat $∣ α - β ∣ = 10$ , then $p$ belongs to the se :