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

Question

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

Tardigrade Admin · Accepted Answer

Given quadratic eqn. is x2+px+(3p/4)=0 So, α+β=-p, αβ=(3p/4) Now, given |α -β |=√10 ⇒ α -β=± √10 ⇒ (α -β ) 2=10 ⇒ α2 +β2-2αβ =10 ⇒ (α +β )2-4αβ =10 ⇒ p2-4×(3p/4)=10 ⇒ p2-3p-10=0 ⇒ p=-2, 5 ⇒ p ∈ -2, 5

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

${2, - 5}$

${- 3, 2}$

${- 2, 5}$

${3, - 5}$

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

{2,−5}

{−3,2}

{−2,5}

{3,−5}

Given quadratic eqn. is x2+px+43p​=0 So, α+β=−p,αβ=43p​ Now, given ∣α−β∣=10​ ⇒α−β=±10​ ⇒(α−β)2=10⇒α2+β2−2αβ=10 ⇒(α+β)2−4αβ=10 ⇒p2−4×43p​=10⇒p2−3p−10=0 ⇒p=−2,5⇒p∈{−2,5}

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

${2, - 5}$

${- 3, 2}$

${- 2, 5}$

${3, - 5}$