Question

Solve by factorisation method √3x² -2√2x - 2√3 = 0

Answer 1

Answer:

• Factorised Form : $(\sqrt{3}x+ \sqrt{2}) (x - \sqrt{6})$

• Roots : $\dfrac{- \sqrt{2}}{\sqrt{3}} \; and \; \sqrt{6}$

Step-by-step explanation:

$\sqrt{3}x^2 - 2\sqrt{2}x - 2\sqrt{3} = 0$

$\implies \sqrt{3}x^2 -3\sqrt{2}x + \sqrt{2}x - 2\sqrt{3} = 0$

$\implies \sqrt{3}x(x - \sqrt{3}\sqrt{2}) +\sqrt{2}(x - \sqrt{3}\sqrt{2}) = 0$

$\implies \sqrt{3}x(x - \sqrt{6}) + \sqrt{2}(x - \sqrt{6}) = 0$

$\implies (\sqrt{3}x+ \sqrt{2}) (x - \sqrt{6}) = 0$

Finding the roots now !

$(i) (\sqrt{3}x+ \sqrt{2}) = 0$

$\implies \sqrt{3}x = -\sqrt{2}$

$\implies x = \dfrac{- \sqrt{2}}{\sqrt{3}}$

$(ii) (x - \sqrt{6}) = 0$

$\implies x = \sqrt{6}$

• Factorised Form : $(\sqrt{3}x+ \sqrt{2}) (x - \sqrt{6})$

• Roots : $\dfrac{- \sqrt{2}}{\sqrt{3}} \; and \; \sqrt{6}$

Thanks !