by completing the square method solve the equation.
x^2-2x-1
can u factor it? when i did it it couldn't be factored. i also don't know why it is not factorable because its a perfect square right?
please help me
It can be factored, but not into integers like (x-1); more like sqrt(2)+-1, and it's definitely not a perfect square. I suspect you've got one of the signs backwards in your head, and you'll probably recognise that as we work through it.
Anyway, direct factorisation wasn't the question; this one is about completing the square.
Forget the constant for a minute, and focus on x^2-2x.
x^2-2x+c is a square. We just have to find c.
We know that
(x+a)^2 = x^2 + 2ax + a^2
and we want
x^2 - 2x + c
so, comparing the coefficients:
2a = -2
a = -1
(x-1)^2 = x^2 - 2x + 1
Our expression is 2 less than that, so:
x^2-2x-1 = (x-1)^2 - 2
To solve the equation using the completing the square method, follow these steps:
Step 1: Move the constant term (in this case, -1) to the right side of the equation:
x^2 - 2x = 1
Step 2: Take half of the coefficient of the x-term (-2) and square it. Add this value to both sides of the equation:
x^2 - 2x + (-2/2)^2 = 1 + (-2/2)^2
x^2 - 2x + 1 = 1 + 1
x^2 - 2x + 1 = 2
Step 3: Rewrite the left side of the equation as a perfect square trinomial. The perfect square trinomial is in the form (x - a)^2, where 'a' is half the coefficient of the x-term.
(x - 1)^2 = 2
Step 4: Take the square root of both sides of the equation:
√((x - 1)^2) = √2
(x - 1) = ±√2
Step 5: Solve for x by adding 1 to both sides of the equation:
x = 1 ±√2
Therefore, the solution to the equation x^2 - 2x - 1 = 0 using the completing the square method is x = 1 ±√2.
Regarding factoring the expression x^2 - 2x - 1, you are correct that it is not factorable in terms of integer values. Even though it is a quadratic equation, it does not factor nicely into two binomials. However, it is still solvable using other methods, such as completing the square or using the quadratic formula.