How do you solve $x^{2} - 5 x - \frac{3}{4} = 0$ $x^2-5x-3/4=0$ by completing the square?

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How do you solve $x^{2} - 5 x - \frac{3}{4} = 0$ $x^2-5x-3/4=0$ by completing the square?

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Answer 1 · 2016-05-13T20:44:17

$x = \frac{5}{2} \pm \sqrt{7}$ $x = 5/2+-sqrt(7)$

Explanation:

Premultiply by $4$ $4$ first to cut down on the fractions involved:

$0 = 4 (x^{2} - 5 x - \frac{3}{4})$ $0 = 4(x^2-5x-3/4)$

$= 4 x^{2} - 20 x - 3$ $=4x^2-20x-3$

$= {(2 x)}^{2} - 2 (2 x) (5) - 3$ $=(2x)^2-2(2x)(5)-3$

$= {(2 x - 5)}^{2} - 5^{2} - 3$ $=(2x-5)^2-5^2-3$

$= {(2 x - 5)}^{2} - 28$ $=(2x-5)^2-28$

$= {(2 x - 5)}^{2} - 2^{2} \cdot 7$ $=(2x-5)^2-2^2*7$

$= {(2 x - 5)}^{2} - {(2 \sqrt{7})}^{2}$ $=(2x-5)^2-(2sqrt(7))^2$

$= ((2 x - 5) - 2 \sqrt{7}) ((2 x - 5) + 2 \sqrt{7})$ $=((2x-5)-2sqrt(7))((2x-5)+2sqrt(7))$

$= (2 x - 5 - 2 \sqrt{7}) (2 x - 5 + 2 \sqrt{7})$ $=(2x-5-2sqrt(7))(2x-5+2sqrt(7))$

$= 4 (x - \frac{5}{2} - \sqrt{7}) (x - \frac{5}{2} + \sqrt{7})$ $=4(x-5/2-sqrt(7))(x-5/2+sqrt(7))$

So $x = \frac{5}{2} \pm \sqrt{7}$ $x = 5/2+-sqrt(7)$

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Explanation:

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