Answer:
[tex]\displaystyle x_1=\frac{5+\sqrt{13}}{2}\approx4.3\text{ and } x_2=\frac{5+\sqrt{13}}{2}\approx0.7[/tex]
Step-by-step explanation:
We are given:
[tex]x^2-5x+3=0[/tex]
And we want to solve this via completing the square.
First, we will subtract 3 from both sides. This yields:
[tex]x^2-5x=-3[/tex]
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:
[tex]\displaystyle x^2-5x+\frac{25}{4}=-3+\frac{25}{4}[/tex]
The left side constitutes a perfect square trinomial as shown:
[tex]\displaystyle (x)^2-2\Big(\frac{5}{2} \Big)(x)+\Big(\frac{5}{2}\Big)^2=-\frac{12}{4}+\frac{25}{4}[/tex]
Factor and subtract:
[tex]\displaystyle \Big(x-\frac{5}{2}\Big)^2=\frac{13}{4}[/tex]
Taking the square root of both sides gives:
[tex]\displaystyle x-\frac{5}{2}=\pm\frac{\sqrt{13}}{2}[/tex]
Hence:
[tex]\displaystyle x=\frac{5}{2}\pm\frac{\sqrt{13}}{2}}=\frac{5\pm\sqrt{13}}{2}[/tex]
Approximate:
[tex]\displaystyle x_1=\frac{5+\sqrt{13}}{2}\approx4.3\text{ and } x_2=\frac{5-\sqrt{13}}{2}\approx0.7[/tex]
