Question

Solve 2x*2+x-4 find the root by completing the square method

Answer 1

Answer:

$<marquee scrollamount=6>pls follow me and mark as brainliest$

Step-by-step explanation:

$\sf\red{ {2x}^{2} + x - 4 = 0 } \\ \sf\red{using \: squere \: method} \\ \sf\purple{ = > \frac{ {2x}^{2} }{2} + \frac{x}{2} + \frac{ - 4}{2} = 0 } \\ \sf\purple{ = > {x}^{2} + \frac{x}{2} - 2 = 0} \\ \sf\blue{ = > {x}^{2} + \frac{1}{4}x + { (\frac{1}{4} }^{2}) = 2+ \frac{1}{16}} \\ \sf\blue{ = > {(x + \frac{1}{4} )}^{2} = \frac{33}{16} } \\ \sf\orange{ = > x + \frac{1}{4} = + - \sqrt{ \frac{33}{4} }} \\ \sf\red{ = > x = + \sqrt{ \frac{33}{16} } - \frac{1}{4} and \: - \sqrt{ \frac{33}{16} } - \frac{1}{4} }$

$\large\boxed{ \fcolorbox{blue}{red}{pls \: follow \: me}}$

$\large\boxed{ \fcolorbox{purple}{yellow}{mark \: as \: brainlist}}$

Answer 2

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{x=\frac{-1\pm\sqrt{33}}{4}}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline \bold{Given :}} \\ \tt: \implies {2x}^{2} + x - 4 = 0 \\ \\ \red{\underline \bold{To \: Find :}} \\ \tt: \implies x = ?$

• According to given question :

$\bold{As \: we \: know \: that} \\ \tt: \implies {2x}^{2} + x - 4 = 0 \\ \\ \text{Both \: side \: dividing \: by \: coefficient \: of \: {x}}^{2} \\ \tt: \implies {x}^{2} + \frac{x}{2} - 2 = 0 \\ \\ \text{both \: side \: adding \: }(\frac{b}{2a} )^{2} \\ \\ \tt \circ \: (\frac{b}{2a} )^{2} = (\frac{ \frac{1}{2} }{2 \times 1} )^{2} = \frac{1}{16} \\ \\ \tt: \implies {x}^{2} + \frac{x}{2} + \frac{1}{16} - 2 = \frac{1}{16} \\ \\ \tt \circ \: {x}^{2} + \frac{x}{2} + \frac{1}{16} = ( {x + \frac{1}{4} })^{2} \\ \\ \tt: \implies ({x + \frac{1}{4} })^{2} - 2 = \frac{1}{16} \\ \\ \tt: \implies (x + \frac{1}{4} )^{2} = \frac{1}{16} + 2 \\ \\ \tt: \implies ( {x + \frac{1}{4} })^{2} = \frac{1 + 32}{16} \\ \\ \tt: \implies ( {x + \frac{1}{4} })^{2} = \frac{33}{16} \\ \\ \tt: \implies x + \frac{1}{4} = \pm\sqrt{ \frac{33}{16} } \\ \\ \tt: \implies x = \pm\frac{ \sqrt{33} }{4} - \frac{1}{4} \\ \\ \green{ \tt: \implies x = \frac{ - 1 \pm \sqrt{33} }{4} }$