1- when a number is squared, result is five times the original number. find the number I did x^2=5x moved the 5x then x^2-5x =0 then?
x = 5
let the number be x
"when a number is squared, result is five times the original number"
----> x^2 = 5x
x^2 - 5x = 0
x(x-5) = 0
so the number is 0 or the number is 5
Well, looks like we have a mathematical mystery to solve! Let's find the number that satisfies the equation x^2 - 5x = 0.
To solve this equation, we want to find the values of x that make the equation true. By factoring out an x, we get x(x - 5) = 0.
So, either x = 0 or (x - 5) = 0. Solving the second equation gives us x = 5.
Therefore, the two possible solutions are x = 0 and x = 5. But remember, we're looking for a number that, when squared, is five times the original number. Squaring 0 gives us 0, which is not five times 0.
Hence, the number you're looking for is x = 5. Squaring 5 does indeed give us 25, which is five times the original number!
So, the number you're searching for is the mighty number 5.
To solve the equation x^2 - 5x = 0, you can factor out an x from both terms:
x(x - 5) = 0
This equation is satisfied when either x = 0 or x - 5 = 0.
For x = 0, when a number is squared, the result is zero, which is not five times the original number.
For x - 5 = 0, solving for x:
x = 5
So, the number you are looking for is x = 5.
To solve the equation x^2 - 5x = 0, you can factor out an x common factor:
x(x - 5) = 0
Now you have two possibilities:
1) Set x = 0:
If x = 0, then the equation becomes 0(0 - 5) = 0, which is true.
2) Set x - 5 = 0:
If x - 5 = 0, then x = 5. Substituting x = 5 into the original equation, we have 5^2 = 5(5), which is also true.
So, the solution to the equation x^2 - 5x = 0 is x = 0 or x = 5.