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How do you solve the quadratic equation by completing the square: $3 x^{2} + 9 x - 1 = 0$ $3x^2+9x-1=0$ ?

1 Answer

Alan P. · Jim H

Jul 16, 2015

$x = - \frac{3}{2} \pm \frac{\sqrt{31}}{2 \sqrt{3}}$ $x = -3/2 +-sqrt(31)/(2sqrt(3))$
$XXXX$ $color(white)("XXXX")$ $XXXX$ $color(white)("XXXX")$ $XXXX$ $color(white)("XXXX")$ (by completing the square)

Explanation:

Given $3 x^{2} + 9 x - 1 = 0$ $3x^2+9x-1=0$

Move the constant to the right side and extract the common factor of $3$ $3$ from the left side:
$XXXX$ $color(white)("XXXX")$ $3 (x^{2} + 3 x) = 1$ $3(x^2+3x) = 1$

If $x^{2} + 3 x$ $x^2+3x$ are the first two terms of the expansion of a squared binomial:
$XXXX$ $color(white)("XXXX")$ ${(x + a)}^{2} = (x^{2} + 2 a x + a^{2})$ $(x+a)^2 = (x^2+2ax+a^2)$

then $a = \frac{3}{2}$ $a= 3/2$ and $a^{2} = {(\frac{3}{2})}^{2} = \frac{9}{4}$ $a^2 = (3/2)^2 = 9/4$
$XXXX$ $color(white)("XXXX")$ which is the amount we will need to add (inside the parentheses) to "complete the square)

$XXXX$ $color(white)("XXXX")$ Note that adding $\frac{9}{4}$ $9/4$ inside the parentheses is the same as adding $3 \cdot (\frac{9}{4})$ $3*(9/4)$ and this amount will need to be added to the right side as well to keep the equation balanced.

$XXXX$ $color(white)("XXXX")$ $3 (x^{2} + 3 x + {(\frac{3}{2})}^{2}) = 1 + \frac{27}{4}$ $3(x^2+3x+(3/2)^2) = 1 + 27/4$

Rewriting as a squared binomial (and simplifying the right side)
$XXXX$ $color(white)("XXXX")$ $3 {(x + \frac{3}{2})}^{2} = \frac{31}{4}$ $3(x+3/2)^2 =31/4$

Dividing both sides by $3$ $3$
$XXXX$ $color(white)("XXXX")$ ${(x + \frac{3}{2})}^{2} = \frac{31}{12}$ $(x+3/2)^2 = 31/12$

Taking the square root of both sides:
$XXXX$ $color(white)("XXXX")$ $x + \frac{3}{2} = \pm \sqrt{\frac{31}{12}}$ $x+3/2 = +- sqrt(31/12)$

Subtracting $3$ $3$ from both sides
$XXXX$ $color(white)("XXXX")$ $x = - \frac{3}{2} \pm \frac{\sqrt{31}}{2 \sqrt{3}}$ $x = -3/2 +-sqrt(31)/(2sqrt(3))$

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