Determine the solution.
x^2=|x|algebraically
I have no idea how to proceed with this. Can you please help?
x²=|x|
means
A. x²=x, or
B. x²=-x
The only solution for case B in the real domain (ℝ) is x=0.
Now let us look at case A.
x²=x
x²-x=0
x(x-1)=0
Can you complete the calculations?
Ok, so I got
x=0 x=1
x=0 x=-1
is that right?
Yes, your solutions are correct.
I have overlooked the root -1 for case B.
(-1)²+(-1)=0
Well done!
Thank you for your guidance! Have a great night.
Thank you, same to you too!
Of course! The equation you have, x^2 = |x|, is a quadratic equation involving an absolute value. To find the solution algebraically, we'll need to consider the cases where |x| is positive and where it is negative.
Let's break down the problem step by step:
1. When x is positive or zero (|x| = x):
In this case, we can rewrite the equation as x^2 = x. Next, we can rearrange the equation to the standard quadratic form, which is x^2 - x = 0. Now, factor out an x from the left side of the equation: x(x - 1) = 0. This equation is satisfied when x = 0 or x - 1 = 0. So, the solutions in this case are x = 0 and x = 1.
2. When x is negative (|x| = -x):
Similarly, we rewrite the equation x^2 = -x. Rearranging it to the standard form gives us x^2 + x = 0. We now factor out an x: x(x + 1) = 0. This equation is satisfied when x = 0 or x + 1 = 0. So, the solutions in this case are x = 0 and x = -1.
To summarize, the solutions to the equation x^2 = |x| are x = -1, 0, and 1.
Please note that we obtained these solutions by considering the possible values for |x|. In the first case, where |x| = x, we found solutions by factoring. In the second case, where |x| = -x, we also found solutions by factoring.