Solve by completing the square.
a² - 2√(3a) + 2 = 0
Solve for x. I don't know if you can use completing the square for this problem.
ax^n + b = 0
suspect you mean
a^2 -2 a sqrt 3 = -2
half of 2 sqrt 3 squared is 3 so add 3 to both sides
a^2 -2 a sqrt 3 + 3 = 1
(a-sqrt 3)^2 = 1
a = sqrt 3 +/-1
To solve the equation \(a^2 - 2\sqrt{3a} + 2 = 0\) by completing the square, follow these steps:
Step 1: Isolate the square term
Move the constant term (2) to the right side of the equation:
\(a^2 - 2\sqrt{3a} = -2\)
Step 2: Complete the square for the variable term
Take half of the coefficient of the square root term (-2\sqrt{3a}), square it, and add it to both sides of the equation:
\(a^2 - 2\sqrt{3a} + (\sqrt{3a})^2 = -2 + (\sqrt{3a})^2\)
\(a^2 - 2\sqrt{3a} + 3a = -2 + 3a\)
Step 3: Simplify
Combine like terms on both sides of the equation:
\(a^2 + a - 2 = 0\)
Now, let's solve for x in the equation \(ax^n + b = 0\):
Step 1: Isolate the variable term
Move the constant term (b) to the right side of the equation:
\(ax^n = -b\)
Step 2: Divide both sides by a
\(\frac{ax^n}{a} = \frac{-b}{a}\)
\(x^n = -\frac{b}{a}\)
Step 3: Take the nth root of both sides
Take the nth root of both sides to solve for x:
\(x = \sqrt[n]{-\frac{b}{a}}\)
Note that the value of n determines the number of possible solutions for x. If n is even, there may be either 0 or 2 solutions (depending on whether \(\frac{b}{a}\) is positive or negative). If n is odd, there will always be 1 solution.