What are the solutions to the equation 1^2=m^2+(-3/2)^2 ?

The solutions to the equation 1^2=m^2+(-3/2)^2 are m=(sqrt(5))/2 i,m=-(sqrt(5))/2 i

Find the zeros of 1^2=m^2+(-3/2)^2

The zeros of 1^2=m^2+(-3/2)^2 are m=(sqrt(5))/2 i,m=-(sqrt(5))/2 i

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1^{2} = m^{2} + (- \frac{3}{2})^{2}

m = \frac{5}{2} i, m = - \frac{5}{2} i

Solution steps

1^{2} = m^{2} + (- \frac{3}{2})^{2}

Expand

1^{2} : 1

Expand

m^{2} + (- \frac{3}{2})^{2} : m^{2} + \frac{9}{4}

1 = m^{2} + \frac{9}{4}

Switch sides

m^{2} + \frac{9}{4} = 1

Move

\frac{9}{4}

to the right side

m^{2} = 1 - \frac{9}{4}

Simplify

1 - \frac{9}{4} : - \frac{5}{4}

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

m = - \frac{5}{4}, m = - - \frac{5}{4}

- \frac{5}{4} = \frac{5}{2} i

- - \frac{5}{4} = - \frac{5}{2} i

m = \frac{5}{2} i, m = - \frac{5}{2} i

\frac{( x - 3 ) ^{2}}{25} + \frac{( 7 - 4 ) ^{2}}{169} = 0

s^{2} + 2 s - 2 = 0

0 = x^{2} + 6 x

x^{2} + x - 4 = 0

5 x^{2} - 2 x - 39 = 0

What are the solutions to the equation 1^2=m^2+(-3/2)^2 ?
The solutions to the equation 1^2=m^2+(-3/2)^2 are m=(sqrt(5))/2 i,m=-(sqrt(5))/2 i
Find the zeros of 1^2=m^2+(-3/2)^2
The zeros of 1^2=m^2+(-3/2)^2 are m=(sqrt(5))/2 i,m=-(sqrt(5))/2 i