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Sum of non real roots of eq.
(x2 + x – 2) · (x2 + x – 3) = 12 is …..
(x2 + x – 2) · (x2 + x – 3) = 12 is …..
Correct answer is '-1'. Can you explain this answer?
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Sum of non real roots of eq.(x2+ x – 2) · (x2+ x – ...
x2 + x = t
t2 – 5t – 6 = 0
t = 6 or t = –1
x2 + x = 6 x2 + x = –1
x2 + x – 6 = 0 x2 + x + 1 = 0
Roots are real sum of roots = –1
t2 – 5t – 6 = 0
t = 6 or t = –1
x2 + x = 6 x2 + x = –1
x2 + x – 6 = 0 x2 + x + 1 = 0
Roots are real sum of roots = –1
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Sum of non real roots of eq.(x2+ x – 2) · (x2+ x – ...
Let's call the equation x^2 + x + k = 0, where k is a constant.
The discriminant of this quadratic equation is Δ = b^2 - 4ac = 1^2 - 4(1)(k) = 1 - 4k.
For the equation to have non-real roots, the discriminant must be negative: Δ < />
So, 1 - 4k < />
=> 4k > 1
=> k > 1/4
Therefore, the non-real roots exist only when k > 1/4.
Since we are asked to find the sum of non-real roots, let's find the roots using the quadratic formula:
x = (-b ± √Δ) / (2a)
For our equation, a = 1, b = 1, and c = k.
x = (-1 ± √(1 - 4k)) / 2
Since we are only interested in the non-real roots, we need to consider the case when the discriminant is negative.
Δ < />
=> 1 - 4k < />
=> 4k > 1
=> k > 1/4
Therefore, the sum of the non-real roots is (-1 + √(1 - 4k))/2 + (-1 - √(1 - 4k))/2 = -1/2.
Hence, the sum of the non-real roots of the equation x^2 + x + k = 0 is -1/2.
The discriminant of this quadratic equation is Δ = b^2 - 4ac = 1^2 - 4(1)(k) = 1 - 4k.
For the equation to have non-real roots, the discriminant must be negative: Δ < />
So, 1 - 4k < />
=> 4k > 1
=> k > 1/4
Therefore, the non-real roots exist only when k > 1/4.
Since we are asked to find the sum of non-real roots, let's find the roots using the quadratic formula:
x = (-b ± √Δ) / (2a)
For our equation, a = 1, b = 1, and c = k.
x = (-1 ± √(1 - 4k)) / 2
Since we are only interested in the non-real roots, we need to consider the case when the discriminant is negative.
Δ < />
=> 1 - 4k < />
=> 4k > 1
=> k > 1/4
Therefore, the sum of the non-real roots is (-1 + √(1 - 4k))/2 + (-1 - √(1 - 4k))/2 = -1/2.
Hence, the sum of the non-real roots of the equation x^2 + x + k = 0 is -1/2.
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Sum of non real roots of eq.(x2+ x – 2) · (x2+ x – 3) = 12 is …..Correct answer is '-1'. Can you explain this answer?
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Sum of non real roots of eq.(x2+ x – 2) · (x2+ x – 3) = 12 is …..Correct answer is '-1'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Sum of non real roots of eq.(x2+ x – 2) · (x2+ x – 3) = 12 is …..Correct answer is '-1'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Sum of non real roots of eq.(x2+ x – 2) · (x2+ x – 3) = 12 is …..Correct answer is '-1'. Can you explain this answer?.
Sum of non real roots of eq.(x2+ x – 2) · (x2+ x – 3) = 12 is …..Correct answer is '-1'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Sum of non real roots of eq.(x2+ x – 2) · (x2+ x – 3) = 12 is …..Correct answer is '-1'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Sum of non real roots of eq.(x2+ x – 2) · (x2+ x – 3) = 12 is …..Correct answer is '-1'. Can you explain this answer?.
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