[tex] \sf{\qquad\qquad\huge\underline{{\sf Answer}}} [/tex]
let's solve ~
[tex]\qquad \sf \dashrightarrow \: {x}^{2} - 18x = 65[/tex]
[tex]\qquad \sf \dashrightarrow \: {x}^{2} - 18x - 65 = 0[/tex]
[tex]\qquad \sf \dashrightarrow \:x {}^{2} - 18x - 65 + 146 = 146[/tex]
[tex]\qquad \sf \dashrightarrow \: {x}^{2} - 18x + 81 = 146[/tex]
[tex]\qquad \sf \dashrightarrow \:(x) {}^{2} - (2 \sdot x \sdot9) + (9) {}^{2} = 146[/tex]
[tex]\qquad \sf \dashrightarrow \:(x - 9) {}^{2} = 146[/tex]
[tex]\qquad \sf \dashrightarrow \:x - 9 = \pm\sqrt{146} [/tex]
[tex]\qquad \sf \dashrightarrow \:x = 9\pm \sqrt{146} [/tex]
Answer:
[tex]x= 9\pm \sqrt{146}[/tex]
Step-by-step explanation:
Completing the Square
When completing the square, the goal is to rewrite the equation to have a perfect square trinomial, which can then be factored.
Step 1:
When completing the square for an equation in the form [tex]x^2+bx+c=0[/tex], the first step is to move the constant to the right side of the equation. This has already been done in the given equation:
[tex]x^2-18x=65[/tex]
Step 2:
Add the square of half the coefficient of x to both sides. This forms a perfect square trinomial on the left side:
[tex]\implies x^2-18x+\left(\dfrac{-18}{2}\right)^2=65+\left(\dfrac{-18}{2}\right)^2[/tex]
Simplify:
[tex]\implies x^2-18x+81=65+81[/tex]
[tex]\implies x^2-18x+81=146[/tex]
Step 3:
Factor the perfect square trinomial on the left side:
[tex]\implies (x-9)^2=146[/tex]
We have now completed the square.
To solve, square root both sides:
[tex]\implies \sqrt{(x-9)^2}=\sqrt{146}[/tex]
[tex]\implies x-9= \pm \sqrt{146}[/tex]
Add 9 to both sides:
[tex]\implies x-9+9= \pm \sqrt{146}+9[/tex]
[tex]\implies x= 9\pm \sqrt{146}[/tex]
Therefore, the solutions of the given equation are:
[tex]x= 9+ \sqrt{146}, \quad x= 9- \sqrt{146}[/tex]
Learn more here:
https://brainly.com/question/27591531
