Question

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

Answer 1

Answer:

$\displaystyle x_1=\frac{5+\sqrt{13}}{2}\approx4.3\text{ and } x_2=\frac{5+\sqrt{13}}{2}\approx0.7$

Step-by-step 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:

$\displaystyle x^2-5x+\frac{25}{4}=-3+\frac{25}{4}$

The left side constitutes a perfect square trinomial as shown:

$\displaystyle (x)^2-2\Big(\frac{5}{2} \Big)(x)+\Big(\frac{5}{2}\Big)^2=-\frac{12}{4}+\frac{25}{4}$

Factor and subtract:

$\displaystyle \Big(x-\frac{5}{2}\Big)^2=\frac{13}{4}$

Taking the square root of both sides gives:

$\displaystyle x-\frac{5}{2}=\pm\frac{\sqrt{13}}{2}$

Hence:

$\displaystyle x=\frac{5}{2}\pm\frac{\sqrt{13}}{2}}=\frac{5\pm\sqrt{13}}{2}$

Approximate:

$\displaystyle x_1=\frac{5+\sqrt{13}}{2}\approx4.3\text{ and } x_2=\frac{5-\sqrt{13}}{2}\approx0.7$