What are the solutions to the equation (3m)^2+(2m+4)^2=(5m)^2 ?

The solutions to the equation (3m)^2+(2m+4)^2=(5m)^2 are m=-2/3 ,m=2

Find the zeros of (3m)^2+(2m+4)^2=(5m)^2

The zeros of (3m)^2+(2m+4)^2=(5m)^2 are m=-2/3 ,m=2

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(3 m)^{2} + (2 m + 4)^{2} = (5 m)^{2}

m = - \frac{2}{3}, m = 2

Decimal Notation

m = - 0.66666 \dots, m = 2

Solution steps

(3 m)^{2} + (2 m + 4)^{2} = (5 m)^{2}

Expand

(3 m)^{2} + (2 m + 4)^{2} : 13 m^{2} + 16 m + 16

Expand

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

13 m^{2} + 16 m + 16 = 25 m^{2}

Move

25 m^{2}

to the left side

- 12 m^{2} + 16 m + 16 = 0

Solve with the quadratic formula

m_{1, 2} = \frac{- 16 \pm 1 6 ^{2} - 4 ( - 12 ) \cdot 16}{2 ( - 12 )}

1 6^{2} - 4 (- 12) \cdot 16 = 32

m_{1, 2} = \frac{- 16 \pm 32}{2 ( - 12 )}

Separate the solutions

m_{1} = \frac{- 16 + 32}{2 ( - 12 )}, m_{2} = \frac{- 16 - 32}{2 ( - 12 )}

m = \frac{- 16 + 32}{2 ( - 12 )} : - \frac{2}{3}

m = \frac{- 16 - 32}{2 ( - 12 )} : 2

The solutions to the quadratic equation are:

m = - \frac{2}{3}, m = 2

563 - \frac{1}{2} 112

16 \cdot \frac{1}{2}

12 x^{2} - 24 x - 96 = 0

∣ 2 x + 1 ∣ \leq 0

15 > - 3 x

What are the solutions to the equation (3m)^2+(2m+4)^2=(5m)^2 ?
The solutions to the equation (3m)^2+(2m+4)^2=(5m)^2 are m=-2/3 ,m=2
Find the zeros of (3m)^2+(2m+4)^2=(5m)^2
The zeros of (3m)^2+(2m+4)^2=(5m)^2 are m=-2/3 ,m=2