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

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

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Answer 1 · 2015-04-30T00:24:50

First, divide by a = 3
$x^{2} - 4 \frac{x}{3} - \frac{5}{3} = 0$ $x^2 - 4x/3 - 5/3 = 0$
$x^{2} - 4 \frac{x}{3} + \frac{4}{9} - \frac{4}{9} - \frac{5}{3} = 0$ $x^2 - 4x/3 + 4/9 - 4/9 - 5/3 = 0$

${(x - \frac{2}{3})}^{2} = \frac{4}{9} + \frac{5}{3} = \frac{19}{9}$ $(x - 2/3)^2 = 4/9 + 5/3 = 19/9$

a. $(x - \frac{2}{3}) = \frac{s q r 19}{3}$ $(x - 2/3) = (sqr19)/3$ ---> $x = \frac{2}{3} + \frac{s q r 19}{3}$ $x = 2/3 + (sqr19)/3$

b. $(x - \frac{2}{3}) = \frac{- s q r 19}{3} \to x = \frac{2}{3} - \frac{s q r 19}{3}$ $(x - 2/3) =(-sqr19)/3 -> x = 2/3 -(sqr19)/3$

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

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