NIMCET 2014 Mathematics PYQ — If and are the roots of the equation , where is a non-zero real n… | Mathem Solvex | Mathem Solvex
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NIMCET 2014 — Mathematics PYQ
NIMCET | Mathematics | 2014
If α and β are the roots of the equation 2x2+2px+p2=0, where p is a non-zero real number, and α4 and β4 are the roots of x2−rx+s=0, then the roots of 2x2−4p2x+4p4−2r=0 are:
Choose the correct answer:
A.
Real and unequal.
B.
Equal and zero.
C.
Imaginary.
(Correct Answer)
D.
Equal and non-zero.
Correct Answer:
Imaginary.
Explanation
Solution
Concept:
The solution to the quadratic equation Ax2+Bx+C=0 can be given by: x=2A−B±B2−4AC
The quantity B2−4AC is called the discriminant (D).
If B2−4AC≥0, the roots are real.
If B2−4AC=0, the roots are real and equal.
If B^2 - 4AC < 0, the roots will be complex and conjugates of each other (imaginary).
For Ax2+Bx+C=0, the sum of roots is −AB and the product of roots is AC.
Calculation:
From the first equation 2x2+2px+p2=0:
α+β=2−2p=−p ---(1)
αβ=2p2 ---(2)
From the second equation x2−rx+s=0:
α4+β4=r ---(3)
Squaring equation (1):
α2+β2+2αβ=p2
Using equation (2):
α2+β2+2(2p2)=p2
⇒α2+β2=0
Squaring again:
(α2+β2)2=02
⇒α4+β4+2α2β2=0
Using equations (2) and (3):
r+2(2p2)2=0
r+2(4p4)=0
r+2p4=0
⇒r=−2p4
---(4)
Now, for the equation 2x2−4p2x+4p4−2r=0, the discriminant is:
D=(−4p2)2−4(2)(4p4−2r)
D=16p4−8(4p4−2r)
D=16p4−32p4+16r
Using equation (4):
D=16p4−32p4+16(−2p4)
D=16p4−32p4−8p4
D=−24p4
Since p is a non-zero real number, p4 is always positive, making the discriminant negative (D < 0).
Therefore, the roots are imaginary.
Correct Option: 3
Explanation
Solution
Concept:
The solution to the quadratic equation Ax2+Bx+C=0 can be given by: x=2A−B±B2−4AC
The quantity B2−4AC is called the discriminant (D).
If B2−4AC≥0, the roots are real.
If B2−4AC=0, the roots are real and equal.
If B^2 - 4AC < 0, the roots will be complex and conjugates of each other (imaginary).
For Ax2+Bx+C=0, the sum of roots is −AB and the product of roots is AC.
Calculation:
From the first equation 2x2+2px+p2=0:
α+β=2−2p=−p ---(1)
αβ=2p2 ---(2)
From the second equation x2−rx+s=0:
α4+β4=r ---(3)
Squaring equation (1):
α2+β2+2αβ=p2
Using equation (2):
α2+β2+2(2p2)=p2
⇒α2+β2=0
Squaring again:
(α2+β2)2=02
⇒α4+β4+2α2β2=0
Using equations (2) and (3):
r+2(2p2)2=0
r+2(4p4)=0
r+2p4=0
⇒r=−2p4
---(4)
Now, for the equation 2x2−4p2x+4p4−2r=0, the discriminant is:
D=(−4p2)2−4(2)(4p4−2r)
D=16p4−8(4p4−2r)
D=16p4−32p4+16r
Using equation (4):
D=16p4−32p4+16(−2p4)
D=16p4−32p4−8p4
D=−24p4
Since p is a non-zero real number, p4 is always positive, making the discriminant negative (D < 0).
