Question

Solve by factorisation method root 3 x^2 -2 root 2x -2 root 3

Answer 1

Step-by-step explanation:

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

$\sqrt{3} {x}^{2} - 3 \sqrt{2} x + \sqrt{2} x - 2 \sqrt{3}$

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

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

Answer 2

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 !