How do you solve using completing the square method $x^{2} + 3 x + 1 = 0$ $x^2+3x+1=0$ ?

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How do you solve using completing the square method $x^{2} + 3 x + 1 = 0$ $x^2+3x+1=0$ ?

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Answer 1 · 2016-03-26T05:43:25

The roots are

$x_{1} = \frac{- 3 + \sqrt{5}}{2}$ $x_1=(-3+sqrt5)/2$ and $x_{1} = \frac{- 3 - \sqrt{5}}{2}$ $x_1=(-3-sqrt5)/2$

Explanation:

From the given $x^{2} + 3 x + 1 = 0$ $x^2+3x+1=0$ , we can see that the coefficient of x^2 is already 1, so we can begin with the coefficient of x which is 3.

The 3 will have to be divided by 2 then the result should be squared and the final result is $\frac{9}{4}$ $color(red)(9/4)$ . This number will be added and subtracted in the equation on one side.

$x^{2} + 3 x + 1 = 0$ $x^2+3x+1=0$

$x^{2} + 3 x + \frac{9}{4} - \frac{9}{4} + 1 = 0$ $x^2+3x+9/4-9/4+1=0$

The first 3 terms now will form a PST-perfect square trinomial.

$x^{2} + 3 x + \frac{9}{4} - \frac{9}{4} + 1 = 0$ $x^2+3x+9/4-9/4+1=0$

$(x^{2} + 3 x + \frac{9}{4}) - \frac{9}{4} + 1 = 0$ $(x^2+3x+9/4)-9/4+1=0$

this $(x^{2} + 3 x + \frac{9}{4})$ $(x^2+3x+9/4)$ is equivalent to ${(x + \frac{3}{2})}^{2}$ $(x+3/2)^2$

So, the equation becomes

${(x + \frac{3}{2})}^{2} - \frac{9}{4} + 1 = 0$ $(x+3/2)^2-9/4+1=0$

simplify

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

transpose the 5/4 to the right side

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

Extract the square root of both sides of the equation

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

$x + \frac{3}{2} = \pm \frac{\sqrt{5}}{2}$ $x+3/2=+-sqrt5/2$

$x = - \frac{3}{2} \pm \frac{\sqrt{5}}{2}$ $x=-3/2+-sqrt5/2$

The roots are

$x_{1} = \frac{- 3 + \sqrt{5}}{2}$ $x_1=(-3+sqrt5)/2$ and $x_{1} = \frac{- 3 - \sqrt{5}}{2}$ $x_1=(-3-sqrt5)/2$

God bless....I hope the explanation is useful.

Answer 2 · 2016-03-26T05:54:34

$x = \frac{\sqrt{5} - 3}{2}$ $x=(sqrt5-3)/2$ or $x = \frac{- \sqrt{5} - 3}{2}$ $x=(-sqrt5-3)/2$

Explanation:

Given equation is $x^{2} + 3 x + 1 = 0$ $x^2+3x+1=0$

$x^{2} + 3 x = - 1$ $x^2+3x=-1$ -----------------------(1)

$T h i r d t e r m = {(\frac{1}{2} \times c o e f f i c i e n t o f x)}^{2}$ $Third term = (1/2 xx coefficient of x)^2$

$T h i r d t e r m = {(\frac{1}{2} \times 3)}^{2}$ $Third term = (1/2 xx 3)^2$

$T h i r d t e r m = {(\frac{3}{2})}^{2}$ $Third term = (3/2)^2$

$T h i r d t e r m = \frac{9}{4}$ $Third term = 9/4$

Add $\frac{9}{4}$ $9/4$ to both sides of equation (1)

$x^{2} + 3 x + \frac{9}{4} = - 1 + \frac{9}{4}$ $x^2+3x+9/4=-1+9/4$

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

$x^{2} + 3 x + \frac{9}{4} = \frac{5}{4}$ $x^2+3x+9/4=5/4$

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

$x + \frac{3}{2} = \sqrt{\frac{5}{4}}$ $x+3/2=sqrt( 5/4)$

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

$x = \pm \frac{\sqrt{5}}{2} - \frac{3}{2}$ $x=+-sqrt5/2-3/2$

$x = \frac{\sqrt{5}}{2} - \frac{3}{2}$ $x=sqrt5/2-3/2$ or $x = - \frac{\sqrt{5}}{2} - \frac{3}{2}$ $x=-sqrt5/2-3/2$

$x = \frac{\sqrt{5} - 3}{2}$ $x=(sqrt5-3)/2$ or $x = \frac{- \sqrt{5} - 3}{2}$ $x=(-sqrt5-3)/2$

$S o l u t i o n s e t = {\frac{\sqrt{5} - 3}{2}, \frac{- \sqrt{5} - 3}{2}}$ $Solution set={(sqrt5-3)/2,(-sqrt5-3)/2}$

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