How do you solve $9 x^{2} + 6 x + 1 = 0$ $9x^2+6x+1=0$ using completing the square?

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How do you solve $9 x^{2} + 6 x + 1 = 0$ $9x^2+6x+1=0$ using completing the square?

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Answer 1 · 2015-06-14T23:58:32

Recognizing that $9 x^{2} + 6 x + 1$ $9x^2+6x+1$ we get $x = - \frac{1}{3}$ $x= -1/3$

Explanation:

$9 x^{2} + 6 x + 1 = 0$ $9x^2+6x+1 = 0$
is equivalent to:
$XXXX$ $color(white)("XXXX")$ ${(3 x + 1)}^{2} = 0$ $(3x+1)^2 = 0$
$\Rightarrow$ $rArr$
$XXXX$ $color(white)("XXXX")$ $3 x + 1 = 0$ $3x+1 = 0$
$\Rightarrow$ $rArr$
$XXXX$ $color(white)("XXXX")$ $x = - \frac{1}{3}$ $x = -1/3$

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How do you solve $9 x^{2} + 6 x + 1 = 0$ $9x^2+6x+1=0$ using completing the square?

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Explanation:

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How do you solve 9x^2+6x+1=09x2+6x+1=09x^2+6x+1=0 using completing the square?

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Explanation:

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How do you solve $9 x^{2} + 6 x + 1 = 0$ $9x^2+6x+1=0$ using completing the square?