solve equation |x^2-2x-6| using quadratic formula. i not get how to do this and it says quadratic formula.
you cannot solve it since you cant square root a negative number... unless -2x was positive 2x.
sorry i missing part of question whole question be |x^2-2x-6| = 4. and answer have to be 1+-sqrt3, 1+-sqrt11 i just not get how to get this.
f(x) = x^2 - 2x - 6 has roots 1±√7 = -1.65,3.65
Between the roots, f(x) < 0 and we have
-(x^2-2x-6) = 4
x^2 - 2x - 2 = 0
x = 1±√3
both those values are in the interval, so they're ok as solutions
Outside that interval,
x^2-2x-6 = 4
x^2 - 2x - 10 = 0
x = 1±√11
both those values are outside the interval, so they're ok as solutions
To solve the equation |x^2-2x-6| = 0 using the quadratic formula, we first need to understand the structure of the given equation.
First, let's remove the absolute value by considering two cases:
Case 1: x^2-2x-6 = 0
In this case, we can directly solve the quadratic equation using the quadratic formula.
The quadratic formula is given by:
x = (-b ± √(b^2-4ac)) / (2a)
For the given equation x^2-2x-6 = 0, we have:
a = 1, b = -2, and c = -6
Plugging these values into the quadratic formula, we get:
x = (-(-2) ± √((-2)^2-4(1)(-6))) / (2(1))
x = (2 ± √(4+24)) / (2)
x = (2 ± √(28)) / (2)
x = (2 ± 2√(7)) / (2)
x = 1 ± √(7)
So, in Case 1, the solutions are x = 1 + √(7) and x = 1 - √(7).
Case 2: -(x^2-2x-6) = 0
In this case, we negate the quadratic equation, -(x^2-2x-6), and solve it in the same way as Case 1.
Negating the equation -(x^2-2x-6) = 0 gives us:
x^2-2x-6 = 0
Since we've already solved this equation in Case 1, we can reuse the solutions: x = 1 + √(7) and x = 1 - √(7).
Therefore, the solutions to the original equation |x^2-2x-6| = 0 are x = 1 + √(7) and x = 1 - √(7).
Note: The absolute value function in the equation |x^2-2x-6| = 0 makes it possible for the equation to have both positive and negative solutions.