Question

solve the equation :
2x⁴-5x²+3=0​

Answer 1

Hello Mate here is your answer below,

$2x {}^{4} -5 x {}^{2} + 3 = 0 \\ = > 2x {}^{4} - 2x {}^{2} - 3x {}^{2} + 3 = 0 \\ = > 2x {}^{2} (x {}^{2} - 1) - 3(x {}^{2} - 1) = 0 \\ = > (2x {}^{2} - 3)(x {}^{2} - 1) = 0 \\ = > 2x {}^{2} - 3 = 0 \\ = > 2x {}^{2} = 3 \\ = > x {}^{2} = \frac{3}{2} \\ = > x = \sqrt{ \frac{3}{2 } } \\ \\ or \\ \\ = > x {}^{2} - 1 = 0 \\ = > x {}^{2} = 1 \\ = > x = \sqrt{1} \\ = > x = 1$

I hope that helps you

Answer 2

SOLUTION

$= > 2x {}^{4} - 5x {}^{2} + 3 = 0 \\ \\ = > 2x {}^{4} - 2x {}^{2} - 3x {}^{2} + 3 = 0 \\ \\ = > 2x {}^{2} (x {}^{2} - 1) - 3( {x}^{2} - 1) = 0 \\ \\ = > (2 {x}^{2} - 3)( {x}^{2} - 1) = 0 \\ \\ = > ( \sqrt{2x} + \sqrt{3} )( \sqrt{2x} - \sqrt{3} )(x + 1)(x - 1) = 0(since \: a {}^{2} - b {}^{2} = (a + b)(a - b) \\ \\ = > x = \frac{ - \sqrt{3} }{ \sqrt{2} } \: or \: x = \frac{ \sqrt{3} }{ \sqrt{2} } \: or \: x = - 1 \: or \: x = 1. \\ \\ = > x = \frac{ - \sqrt{6} }{2} \: or \: x = \frac{ \sqrt{6} }{2} \: or \: x = - 1 \: or \: x = 1. \: answer$

Hope it helps