# What is m is 1/m^2 = 1/m - 1/2

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Given the equation 1/m^2 = 1/m - 1/2.

We need to find the values of m that satisfies the equation.

Let us get rid of the denominator by multiplying by 2m^2.

==> 2m^2(1/m^2) = 2m^2(1/m) - 2m^2(1/2)

==> 2 = 2m - m^2

Now we will combine all terms on the left side.

==> m^2 - 2m + 2 = 0.

Now we will use the formula to solve for m.

==> m1= ( 2 + sqrt(4-8) / 2

= (2+sqrt-4)/2

= (2 + 2i)/2

= 1+ i

==> m2= 1-i

Then there are **no real** solution for m.

However, the equation has a complex solution.

**==> m = { 1+i, 1-i}**

The values of m that verify the equation arethe roots of the equation.

1/m^2 = 1/m - 1/2

Since the LCD is 2m^2, we'll multiply by 2m^2 both sides.

2m^2/m^2 = 2m^2/m - 2m^2/2

We'll simplify and we'll get:

2 = 2m - m^2

We'll move all terms to one side:

m^2 - 2m + 2 = 0.

We'll re-write the equation:

m^2 - 2m + 1 + 1 = 0

(m-1)^2 = -1

We'll have square root both sides:

sqrt (m-1)^2 = sqrt -1

m-1 = -i

m1 = 1 - i

m-1 = +i

m2 = 1 + i

**The solutions of the equation are :{1 - i ; 1 + i}.**