Скачать или смотреть For real number x and y, if x2 + xy -y2 + 2x-y + 1 = 0, then :

For real number x and y, if x2 + xyy2 + 2xy + 1 = 0, then :
(1)y can not be between 0 and 85
(2)y can not be between − 85 and 85
(4)y can not be between − 16 and 0 and 0
Correct Answer: Option 3
(3)y can not be between − 8

JEE Math Trick: Quadratic Equation with Logarithmic Coefficients | Find \alpha^2 + \alpha\beta + \beta^2"
This video breaks down a challenging quadratic equation problem that often appears in competitive exams like JEE Main and Advanced. 🚀
We tackle the problem of finding \alpha^2 + \alpha\beta + \beta^2 where \alpha and \beta are the roots of the equation:
x^2 - \left(3 + 2^{\sqrt{\log_2 3}} - 3^{\sqrt{\log_3 2}}\right)x - 2\left(3^{\log_3 2} - 2^{\log_2 3}\right) = 0
What you will learn:
• Logarithm Simplification: Learn key properties like a^{\log_a b} = b and the change of base rule \log_a b = \frac{1}{\log_b a} to simplify the complex coefficients.
• Vieta's Formulas: Review how to use the sum of roots (\alpha + \beta) and product of roots (\alpha\beta) to solve for symmetric expressions.
• Algebraic Identity: Master the identity \alpha^2 + \alpha\beta + \beta^2 = (\alpha + \beta)^2 - \alpha\beta.
Watch till the end to see the full, step-by-step solution and arrive at the correct multiple-choice answer!
#JEEMain #QuadraticEquations #Logarithms #MathTricks #Algebra #competitiveexams


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