Answer:
[tex]x=\dfrac{3 \pm \sqrt{29}}{2}[/tex]
Step-by-step explanation:
Given equation:
[tex]x^2-3x-5=0[/tex]
To find the exact solutions of the given equation, use the quadratic formula or complete the square.
Method 1: Quadratic Formula
[tex]x=\dfrac{-b \pm \sqrt{b^2-4ac} }{2a}\quad\textsf{when }\:ax^2+bx+c=0[/tex]
Therefore:
[tex]a=1, \quad b=-3, \quad c=-5[/tex]
Substituting these values into the formula:
[tex]\implies x=\dfrac{-(-3) \pm \sqrt{(-3)^2-4(1)(-5)} }{2(1)}[/tex]
[tex]\implies x=\dfrac{3 \pm \sqrt{29}}{2}[/tex]
Method 2: Completing the square
Move the constant to the right side by adding 5 to both sides:
[tex]\implies x^2-3x-5+5=0+5[/tex]
[tex]\implies x^2-3x=5[/tex]
Add the square of half the coefficient of x to both sides:
[tex]\implies x^2-3x+\left(\dfrac{-3}{2}\right)^2=5+\left(\dfrac{-3}{2}\right)^2[/tex]
[tex]\implies x^2-3x+\dfrac{9}{4}=\dfrac{29}{4}[/tex]
Factor the perfect trinomial on the left side:
[tex]\implies \left(x-\dfrac{3}{2}\right)^2=\dfrac{29}{4}[/tex]
Square root both sides:
[tex]\implies \sqrt{\left(x-\dfrac{3}{2}\right)^2}=\sqrt{\dfrac{29}{4}}[/tex]
[tex]\implies x-\dfrac{3}{2}=\pm\dfrac{\sqrt{29}}{2}[/tex]
Add 3/2 to both sides:
[tex]\implies x-\dfrac{3}{2}+\dfrac{3}{2}=\pm\dfrac{\sqrt{29}}{2}+\dfrac{3}{2}[/tex]
[tex]\implies x=\dfrac{3 \pm \sqrt{29}}{2}[/tex]
Learn more about completing the square here:
https://brainly.com/question/27933930
