Question

PLEASE HELP! Using completing the square to solve x2-5x+3=0. Round to the nearest tenth as needed

Answer 1

`\displaystyle x_1=(5+√(13))/(2)\approx4.3\text{ and } x_2=(5+√(13))/(2)\approx0.7`

Explanation:

We are given:

`x^2-5x+3=0`

And we want to solve this via completing the square.

First, we will subtract 3 from both sides. This yields:

`x^2-5x=-3`

Next, we will factor out the leading coefficient from the left.

In this case, the leading coefficient is simply 1 so we can leave it as is.

Next, we want to add a constant that is half of the b coefficient squared.

In this case, b = -5.

Half of that is -5/2.

Squaring yields 25/4.

So, we will add 25/4 to both sides. Since our leading coefficient is 1, we don't need to multiply by anything when we add it on the right. Hence:

`\displaystyle x^2-5x+(25)/(4)=-3+(25)/(4)`

The left side constitutes a perfect square trinomial as shown:

`\displaystyle (x)^2-2\Big((5)/(2) \Big)(x)+\Big((5)/(2)\Big)^2=-(12)/(4)+(25)/(4)`

Factor and subtract:

`\displaystyle \Big(x-(5)/(2)\Big)^2=(13)/(4)`

Taking the square root of both sides gives:

`\displaystyle x-(5)/(2)=\pm(√(13))/(2)`

Hence:

`\displaystyle x=(5)/(2)\pm(√(13))/(2)}=(5\pm√(13))/(2)`

Approximate:

`\displaystyle x_1=(5+√(13))/(2)\approx4.3\text{ and } x_2=(5-√(13))/(2)\approx0.7`