What are the solutions to the equation (2m+2)^2+(3m)^2=(5m)^2 ?

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

Find the zeros of (2m+2)^2+(3m)^2=(5m)^2

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

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(2 m + 2)^{2} + (3 m)^{2} = (5 m)^{2}

m = - \frac{1}{3}, m = 1

Solution steps

(2 m + 2)^{2} + (3 m)^{2} = (5 m)^{2}

Expand

(2 m + 2)^{2} + (3 m)^{2} : 13 m^{2} + 8 m + 4

Expand

(5 m)^{2} : 25 m^{2}

13 m^{2} + 8 m + 4 = 25 m^{2}

Move

25 m^{2}

to the left side

- 12 m^{2} + 8 m + 4 = 0

Solve with the quadratic formula

m_{1, 2} = \frac{- 8 \pm 8 ^{2} - 4 ( - 12 ) \cdot 4}{2 ( - 12 )}

8^{2} - 4 (- 12) \cdot 4 = 16

m_{1, 2} = \frac{- 8 \pm 16}{2 ( - 12 )}

Separate the solutions

m_{1} = \frac{- 8 + 16}{2 ( - 12 )}, m_{2} = \frac{- 8 - 16}{2 ( - 12 )}

m = \frac{- 8 + 16}{2 ( - 12 )} : - \frac{1}{3}

m = \frac{- 8 - 16}{2 ( - 12 )} : 1

The solutions to the quadratic equation are:

m = - \frac{1}{3}, m = 1

x^{2} + 16 x - 145 = 10 x - 7

3.4 = - 13.6 + (- 3.4 c) + 1.7 c

x^{2} + 21 < 10 x

x^{2} = 7 x + 3

5 y - [2 (y + 1) + 2] = 7 y

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