Question

Solve 2x^2+x-4=0 by completing the squares method

Answer 1

${2x}^{2} + x - 4 = 0 \\ {2x}^{2} + x = 4 \\ divide \: both \: sides \: by \: {x}^{2} coefficient \\ \frac{ {2x}^{2} }{2} + \frac{x}{2} = \frac{4}{2} \\ {x}^{2} + \frac{x}{2} = 2 \\ {x}^{2} + 2.x. \frac{1}{4} = 2 \\ add \: ( \frac{1}{4} ) {}^{2} \: on \: both \: sides \\ {x}^{2} + 2.x. \frac{1}{4} + ( \frac{1}{4} ) {}^{2} = 2 + ( \frac{1}{4} ) {}^{2} \\ l.h.s. \: is \: in \: the \: form \: of \: (a + b) { }^{2} \\ (a + b) {}^{2} = a {}^{2} + 2ab + b {}^{2} \\ (x + \frac{1}{4} ) {}^{2} = 2 + \frac{1}{16} \\ (x + \frac{1}{4} ) {}^{2} = \frac{33}{16} \\ x + \frac{1}{4} = \sqrt{ \frac{33}{16} } \\ x + \frac{1}{4} = + or - \frac{ \sqrt{33} }{4} \\ x = \frac{ \sqrt{33} }{4} - \frac{1}{4} \\ x= \frac{ \sqrt{33} - 1 }{4} \\ also \: \\ x = \frac{ - \sqrt{33} }{4} - \frac{1}{4} \\ x = \frac{ - \sqrt{33} - 1 }{4} \\take \: minus \: as \: common \\ x = \frac{ - ( \sqrt{33} + 1)}{4}$

Hope this helps you mate........

Please mark it as brainliest.........!!!!!!!!