If $3 x^{2} - 2 x + 7 = 0$ $3x^2-2x+7=0$ , what is ${(x - \frac{1}{3})}^{2}$ $(x-1/3)^2$ ?

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If $3 x^{2} - 2 x + 7 = 0$ $3x^2-2x+7=0$ , what is ${(x - \frac{1}{3})}^{2}$ $(x-1/3)^2$ ?

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Answer 1 · 2016-07-30T23:32:35

not sure what you're needing here

Explanation:

is this a substitution question?

Answer 2 · 2016-07-30T23:37:25

${(x - \frac{1}{3})}^{2} = - \frac{20}{9}$ $(x-1/3)^2 = -20/9$

Explanation:

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

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

$= 3 ({(x - \frac{1}{3})}^{2} - \frac{1}{9} + \frac{7}{3})$ $= 3((x-1/3)^2-1/9+7/3)$

$= 3 ({(x - \frac{1}{3})}^{2} + \frac{20}{9})$ $= 3((x-1/3)^2+20/9)$

Hence ${(x - \frac{1}{3})}^{2} = - \frac{20}{9}$ $(x-1/3)^2 = -20/9$

So $x - \frac{1}{3} = \pm \sqrt{- \frac{20}{9}} = \pm \frac{2 \sqrt{5}}{3} i$ $x-1/3 = +-sqrt(-20/9) = +-(2sqrt(5))/3i$

So $x = \frac{1}{3} \pm \frac{2 \sqrt{5}}{3} i$ $x=1/3+-(2sqrt(5))/3i$

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If $3 x^{2} - 2 x + 7 = 0$ $3x^2-2x+7=0$ , what is ${(x - \frac{1}{3})}^{2}$ $(x-1/3)^2$ ?

2 Answers

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