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The roots of the equation x2 + (2p–1)x + p2 = 0 are real if.
- a)p > 1
- b)p < 4
- c)p > 1/4
- d)p < 1/4
Correct answer is option 'D'. Can you explain this answer?
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The roots of the equation x2 + (2p–1)x + p2 = 0 are real if.a)p ...
Given equation: x^2 + (2p+1)x + p^2 = 0
To find: The condition for roots to be real.
Solution:
For the roots of a quadratic equation to be real, the discriminant of the quadratic equation should be greater than or equal to zero.
Discriminant (D) = b^2 - 4ac
Here, a = 1, b = 2p+1, c = p^2
Substituting the values, we get:
D = (2p+1)^2 - 4(1)(p^2)
D = 4p^2 + 4p + 1 - 4p^2
D = 4p + 1
To find the condition for real roots, we equate D to zero.
4p + 1 = 0
4p = -1
p = -1/4
Therefore, the roots of the given equation will be real if p = -1/4.
Hence, the correct answer is option D.
To find: The condition for roots to be real.
Solution:
For the roots of a quadratic equation to be real, the discriminant of the quadratic equation should be greater than or equal to zero.
Discriminant (D) = b^2 - 4ac
Here, a = 1, b = 2p+1, c = p^2
Substituting the values, we get:
D = (2p+1)^2 - 4(1)(p^2)
D = 4p^2 + 4p + 1 - 4p^2
D = 4p + 1
To find the condition for real roots, we equate D to zero.
4p + 1 = 0
4p = -1
p = -1/4
Therefore, the roots of the given equation will be real if p = -1/4.
Hence, the correct answer is option D.
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The roots of the equation x2 + (2p–1)x + p2 = 0 are real if.a)p > 1b)p < 4c)p > 1/4d)p < 1/4Correct answer is option 'D'. Can you explain this answer?
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