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If a, b, c are real, then both the roots of the equation (x – b) (x – c) + (x – c) (x – a) + (x – a) (x – b) = 0 are always
- a)positive
- b)negative
- c)real
- d)imaginary
Correct answer is option 'C'. Can you explain this answer?
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If a, b, c are real, then both the roots of the equation (x – b)...
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If a, b, c are real, then both the roots of the equation (x – b)...
We need to prove that both the roots of the given equation are positive.
Let's first find the discriminant of the given equation:
D = b^2 - 4ac
= (a + b + c)^2 - 4ac
= a^2 + b^2 + c^2 + 2ab + 2ac + 2bc - 4ac
= a^2 + b^2 + c^2 - 2ac + 2ab + 2bc
= (a - c)^2 + b^2 + 2ab + 2bc
Since a, b, and c are real, we know that D ≥ 0 for the given equation to have real roots.
Now, let's find the roots using the quadratic formula:
x = (-b ± √D) / 2a
= (-(a + b + c) ± √[(a - c)^2 + b^2 + 2ab + 2bc]) / 2a
Let's consider the positive root:
x = (-(a + b + c) + √[(a - c)^2 + b^2 + 2ab + 2bc]) / 2a
= (-a - b - c + √[(a - c)^2 + b^2 + 2ab + 2bc]) / 2a
Since D ≥ 0, we know that the term inside the square root is non-negative. Therefore,
(a - c)^2 + b^2 + 2ab + 2bc ≥ 0
a^2 - 2ac + c^2 + b^2 + 2ab + 2bc ≥ 0
(a + b)^2 + (b + c)^2 + (c - a)^2 ≥ 0
This inequality holds true for all real values of a, b, and c. Therefore,
(-(a + b + c) + √[(a - c)^2 + b^2 + 2ab + 2bc]) / 2a > 0
x > 0
Similarly, we can prove that the negative root is also positive:
x = (-(a + b + c) - √[(a - c)^2 + b^2 + 2ab + 2bc]) / 2a
= (-a - b - c - √[(a - c)^2 + b^2 + 2ab + 2bc]) / 2a
(-(a + b + c) - √[(a - c)^2 + b^2 + 2ab + 2bc]) / 2a < 0="" />
x < 0="" />
Therefore, both roots of the given equation are positive.
Let's first find the discriminant of the given equation:
D = b^2 - 4ac
= (a + b + c)^2 - 4ac
= a^2 + b^2 + c^2 + 2ab + 2ac + 2bc - 4ac
= a^2 + b^2 + c^2 - 2ac + 2ab + 2bc
= (a - c)^2 + b^2 + 2ab + 2bc
Since a, b, and c are real, we know that D ≥ 0 for the given equation to have real roots.
Now, let's find the roots using the quadratic formula:
x = (-b ± √D) / 2a
= (-(a + b + c) ± √[(a - c)^2 + b^2 + 2ab + 2bc]) / 2a
Let's consider the positive root:
x = (-(a + b + c) + √[(a - c)^2 + b^2 + 2ab + 2bc]) / 2a
= (-a - b - c + √[(a - c)^2 + b^2 + 2ab + 2bc]) / 2a
Since D ≥ 0, we know that the term inside the square root is non-negative. Therefore,
(a - c)^2 + b^2 + 2ab + 2bc ≥ 0
a^2 - 2ac + c^2 + b^2 + 2ab + 2bc ≥ 0
(a + b)^2 + (b + c)^2 + (c - a)^2 ≥ 0
This inequality holds true for all real values of a, b, and c. Therefore,
(-(a + b + c) + √[(a - c)^2 + b^2 + 2ab + 2bc]) / 2a > 0
x > 0
Similarly, we can prove that the negative root is also positive:
x = (-(a + b + c) - √[(a - c)^2 + b^2 + 2ab + 2bc]) / 2a
= (-a - b - c - √[(a - c)^2 + b^2 + 2ab + 2bc]) / 2a
(-(a + b + c) - √[(a - c)^2 + b^2 + 2ab + 2bc]) / 2a < 0="" />
x < 0="" />
Therefore, both roots of the given equation are positive.
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If a, b, c are real, then both the roots of the equation (x – b) (x – c) + (x – c) (x – a) + (x – a) (x – b) = 0 are alwaysa)positiveb)negativec)reald)imaginaryCorrect answer is option 'C'. Can you explain this answer?
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