How do you solve $2 x^{2} - 7 x + 3 = 0$ $2x^2 -7x+3=0$ by completing the square?

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Answer 1 · 2017-08-17T14:33:30

$x = 3$ $x=3$ or $x = \frac{1}{2}$ $x=1/2$

$2 x^{2} - 7 x + 3 = 0$ $2x^2-7x+3=0$

$\Rightarrow 2 (x^{2} - \frac{7}{2} x) + 3 = 0$ $implies 2(x^2-7/2 x)+3=0$

$\Rightarrow 2 (x^{2} - 2 \times \frac{7}{4} x + \frac{49}{16}) = \frac{49}{8} - 3 = \frac{25}{8}$ $implies 2(x^2-2 times 7/4 x+49/16)=49/8-3 = 25/8$

${(x - \frac{7}{4})}^{2} = \frac{25}{16} \Rightarrow x - \frac{7}{4} = \pm \frac{5}{4}$ $(x-7/4)^2 = 25/16 implies x-7/4 = pm 5/4$

$\Rightarrow x = \frac{7}{4} \pm \frac{5}{4} = 3 or \frac{1}{2}$ $implies x=7/4 pm 5/4 = 3 or 1/2$

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