Given that 𝑥 is contained in all reals and negative 𝑥 over 10 is equal to 100 over 𝑥 squared, determine the value of 𝑥.
The first thing I’m gonna do is just copy down this equation. To solve for 𝑥, I need to get both of my 𝑥s on the same side of the equation. And I’ll need them to be out of the denominator. To fix this, we can multiply the right side of the equation by 𝑥 squared over one. And if we multiply the right side of the equation by 𝑥 squared over one, we have to multiply the left side of the equation by 𝑥 squared over one.
On the left, we’ll have 𝑥 squared times negative 𝑥 over 10. On the right, the two 𝑥 squareds cancel out, leaving us with 100. I can multiply 𝑥 squared times negative 𝑥. It equals negative 𝑥 cubed. And we’ll copy everything else down. Again, we’re trying to isolate 𝑥. We need to get the number 10 out of the denominator and onto the other side of the equation. So I multiply by 10 over one on both sides. On the left, we’re left with negative 𝑥 cubed equals 1000.
I noticed that I have a negative 𝑥-value. So I want to multiply both sides of this equation by negative one. After I do that, I have an equation that says 𝑥 cubed equals negative 1000. After that, we need to take the cube root of both sides of the equation. The cube root of 𝑥 cubed equals 𝑥. And the cube root of negative 1000 equals negative 10.
Let’s check and see if this makes sense in the context of our original equation. Negative negative 10 over 10, which is 10 over 10, is equal to 100 over negative 10 squared. Negative 10 squared is 100. 10 over 10 is equal to 100 over 100. So it seems we have a valid 𝑥-value. The value for 𝑥 that makes this statement true is negative 10.