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

##### 2 Answers

The roots are

${x}_{1} = \frac{- 3 + \sqrt{5}}{2}$ and ${x}_{1} = \frac{- 3 - \sqrt{5}}{2}$

#### Explanation:

From the given ${x}^{2} + 3 x + 1 = 0$, 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 $\textcolor{red}{\frac{9}{4}}$. This number will be added and subtracted in the equation on one side.

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

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

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

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

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

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

So, the equation becomes

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

simplify

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

transpose the 5/4 to the right side

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

Extract the square root of both sides of the equation

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

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

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

The roots are

${x}_{1} = \frac{- 3 + \sqrt{5}}{2}$ and ${x}_{1} = \frac{- 3 - \sqrt{5}}{2}$

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

Mar 26, 2016

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

#### Explanation:

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

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

$T h i r d t e r m = {\left(\frac{1}{2} \times c o e f f i c i e n t o f x\right)}^{2}$

$T h i r d t e r m = {\left(\frac{1}{2} \times 3\right)}^{2}$

$T h i r d t e r m = {\left(\frac{3}{2}\right)}^{2}$

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

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

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

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

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

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

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

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

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

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

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

$S o l u t i o n s e t = \left\{\frac{\sqrt{5} - 3}{2} , \frac{- \sqrt{5} - 3}{2}\right\}$