If (2\sqrt2x^3+3\sqrt3y^3)\div(\sqrt2x+\sqrt3y) =(\A x^2+\B xy+\C y^2), then (2\A-\sqrt6\B-4\C)=?

SSC CGL Model Paper 11

Question : 52 of 100

Marks: +1, -0

(2√2x^3+3√3y^3)÷(√2x+√3y)

=(\A x^2+\B xy+\C y^2)

, then

(2\A-√6\B-4\C)=?

Solution:

{( √2x)^3+ (√3y)^3}/{√2x+√3y}

=(\A x^2+\B xy+\C y^2)

⇒2x^2+3y^2-√6xy

=\A x^2+\B xy+\C y^2

Compare on both side.

\A=2,\C=3,\B=-√6

∴2\A-√6\B-4\C

=4-√6(-√6)-4×3=-2

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