Solve the equation |x^2 - 2x - 6| = 4 algebraiclly, using the quadratic formula
|x^2 - 2x - 6| = 4
so
case 1: x^2 - 2x - 6 = 4 OR case 2: x^2 - 2x - 6 = - 4
I will do case 1, leaving it up to you to solve case 2 in the same way
x^2 - 2x - 6 = 4
x^2 - 2x - 10 = 0
x = (2 ± √44)/2
= (2 ± 2√11)/2
= 1 ± √11
To solve the equation |x^2 - 2x - 6| = 4 algebraically, first we need to consider two cases: when the expression inside the absolute value is positive and when it is negative. Let's solve for each case separately.
Case 1: x^2 - 2x - 6 = 4
To solve this equation algebraically, we need to move all terms to one side of the equation to obtain a quadratic equation in standard form.
x^2 - 2x - 6 - 4 = 0
x^2 - 2x - 10 = 0
Now, we can apply the quadratic formula to solve for x:
x = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = 1, b = -2, and c = -10. Substituting these values into the quadratic formula:
x = (-(-2) ± √((-2)^2 - 4(1)(-10))) / (2(1))
x = (2 ± √(4 + 40)) / 2
x = (2 ± √44) / 2
x = (2 ± 2√11) / 2
x = 1 ± √11
So, the solutions for this case are x = 1 + √11 and x = 1 - √11.
Case 2: -x^2 + 2x + 6 = 4
Similar to Case 1, we move all terms to one side of the equation:
-x^2 + 2x + 6 - 4 = 0
-x^2 + 2x + 2 = 0
Again, we can use the quadratic formula to solve for x:
x = (-b ± √(b^2 - 4ac)) / (2a)
Here, a = -1, b = 2, and c = 2. Substituting these values into the quadratic formula:
x = (-(2) ± √((2)^2 - 4(-1)(2))) / (2(-1))
x = (-2 ± √(4 + 8)) / -2
x = (-2 ± √12) / -2
x = (-2 ± √(4 * 3)) / -2
x = (-2 ± 2√3) / -2
x = 1 ± √3
Thus, the solutions for this case are x = 1 + √3 and x = 1 - √3.
Hence, the complete solution to the equation |x^2 - 2x - 6| = 4 is x = 1 + √11, x = 1 - √11, x = 1 + √3, and x = 1 - √3.