Given that x- y - R- solve 4x2 - 3xy - 2xy - 3x2i - 4y2 - x2-2 - 3xy - 2y2 i

In the given equation, real parts of both LHS and RHS will be equal #160;Therefore, 4x^2+3xy=4y^2 x^229x^2+6xy 8y^2=0 #160;9x^2+12xy 6xy 8y^2=0 ...

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Finding Square of a Number Using Identity

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Question

Given that $x, y ϵ R$ , solve $4 x^{2} + 3 x y + (2 x y - 3 x^{2}) i = 4 y^{2} - (\frac{x^{2}}{2}) + (3 x y - 2 y^{2}) i$

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x, y ϵ R

4 x^{2} + 3 x y + (2 x y - 3 x^{2}) i = 4 y^{2} - (\frac{x^{2}}{2}) + (3 x y - 2 y^{2}) i

x

y

(2 + 3 i) x^{2} - (3 - 2 i) y = 2 x - 3 y + 5 i

4 x^{2} + 3 x y + (2 x y - 3 x^{2}) i = 4 y^{2} - (\frac{x^{2}}{2}) + (3 x y - 2 y^{2}) i

Simplify $(- 4 x^{2} - 2 x y + 9 y^{2}) - (3 x^{2} - 3 x y + 7 y^{2})$

3 x^{2} + 5 x y - 4 y^{2} + x^{2} - 8 x y - 5 y^{2}

4 x^{2} - 3 x y - 9 y^{2}

(2 x y - x y) (3 x y - 5)

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