how to solve by factorization and can you explain? x^2= 6x
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You need to move all terms that contain x to one side such that:
`x^2 - 6x = 0 `
Notice that both terms have a common factor x, hence, factoring out x yields:
`x(x - 6) = 0`
Since a product of two factors is 0 when a factor or both are zeroes, hence, you need to set the following equations such that:
`{(x=0),(x-6=0):} =gt {(x=0),(x=6):}`
Hence, evaluating the solution to the given equation using factorization, yields `x = 0` and `x = 6.` `.`
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The equation x^2= 6x has to be solved.
x^2 = 6x
Subtract 6x from both the sides
x^2 - 6x = 6x - 6x
x^2 - 6x = 0
Isolate the common factor x.
x(x - 6) = 0
x - 6 = 0 gives x = 6
The solution of the equation x^2 = 6x is x = 0 and x = 6
In order to solve bu factorization, you want to get each equation in a form where the right-hand side equals zero. So for the first case: x^2 = 6x is equivalent to x^2 - 6x =0.
We then try to factorize the left-hand side. Both terms involve x so we can factorize that out: x^2 - 6x = x*(x - 6)
Hence x*(x-6) = 0. Now, because we get zero whenever we multiply something by zero, the equation implies that either x=0 or x-6=0. Hence x=0 or 6.
For the second equation: 2x^2 = x/3 => 2x^2 - x/3 = 0 => 6x^2 - x =0.
As before we can see that x is a common factor so we get
6x^2 - x = x*(6x - 1)
So either x =0 or 6x - 1 =0 => 6x=1 => x=1/6. So x= 0 or 1/6.
For the third equation: x/5 - x^2 = 0 => x - 5*x^2 = 0 => x*(1 - 5x) = 0 => Either x=0 or 1-5x=0
For 1 - 5*x = 0 => 5*x = 1 => x = 1/5
hence either x=0 or x=1/5.
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