If α, β are the real and distinct roots of x2 + px + q = 0 and α4, β4 are the roots of x2 – rx + s = 0, then the equation x2 − 4qx + 2q2 − r = 0 has always

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Two real roots

Two negative roots

Two positive roots

One positive and one negative root

answer is D.

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Detailed Solution

Given, the real and distinct roots equation x2 + px + q = 0 are α, β and α4, β4 are the roots of x2 – rx + s = 0
Since we know that in an equation ax2 + bx + c, the sum of roots of the equation is equal −a
and product is equal to b.
Using this in the equation x2 + px + q = 0, we get
⇒ α + β = −p
⇒ αβ = q
Now, considering the equation x2 – rx + s = 0, we get
⇒ α4+β4 = r ......eq. (1)
⇒ α4β4 = s ......eq. (2)
Adding and subtracting the left side of the equation (1) by 2α2β2, we get
⇒ α4 + β4 + 2α2β2 − 2α2β2 = r
Using the property, (a+b)2 = a2 + b2 + 2ab in the above equation, we get
⇒ (α2+β2)2 − 2αβ = r
Adding and subtracting the left side of the above equation by 2αβ in the bracket, we get
⇒ (α2+β2+2αβ−2αβ)2 − 2α2β2 = r
Using the property, (a+b)2 = a2 + b2 + 2ab in the above equation, we get
⇒ ((α+β)2 − 2αβ)2 − 2αβ = r
Substituting the value of α + β and αβ in the above equation, we get
⇒ ((−p)2−2q)2−2q2 = r
⇒ (p2−2q)2−2q2 = r
⇒ p4 + 4q2 − 4p2q − 2q2 = r
⇒ p4 + 2q2 − 4p2q = r
Rearranging the above equation, we get
⇒ 2q2 − r = 4p2q−p4
⇒ 2q2 – r = p2(4q−p2)
Since we know that p2 > 0 and 4q − p2 < 0, so the above equation will become
⇒ 2q2 – r < 0
Since a product of the roots can only be negative if one root is negative and one root is positive.
Using the above equation, we get that x2 − 4qx + 2q2 − r = 0 has one positive and one negative root,
Hence, option 4 is correct.

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