207k views
5 votes
Solve for all values of x x^2-2x-48=0
by completing the square

User Essayoub
by
7.3k points

1 Answer

2 votes

Answer: To solve for all values of x in the equation x^2 - 2x - 48 = 0 by completing the square, we can follow these steps:

Bring the x term to one side of the equation by adding 2x to both sides: x^2 - 2x = 48

Divide the coefficient of the x^2 term (which is 1) by 2 and square it to find the value to add to both sides of the equation to complete the square: (1/2)^2 = 1/4. Add 1/4 to both sides of the equation: x^2 - 2x + 1/4 = 48 + 1/4

Take the square root of both sides of the equation: (x - 1)^2 = 192 + 1

Solve for x by taking the square root of both sides of the equation: x - 1 = ± √193

Add 1 to both sides of the equation to find the values of x: x = 1 ± √193

So the solutions of the equation are x = 1 + √193 and x = 1 - √193

It is important to notice that these are the solutions in radical form, so they are not exact values of x.

Explanation:

User Thibault Witzig
by
6.7k points