What are the solutions to the equation (3x)^2+(4x)^2=75^2 ?

The solutions to the equation (3x)^2+(4x)^2=75^2 are x=15,x=-15

Find the zeros of (3x)^2+(4x)^2=75^2

The zeros of (3x)^2+(4x)^2=75^2 are x=15,x=-15

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(3 x)^{2} + (4 x)^{2} = 7 5^{2}

x = 15, x = - 15

Solution steps

(3 x)^{2} + (4 x)^{2} = 7 5^{2}

Expand

(3 x)^{2} + (4 x)^{2} : 25 x^{2}

Expand

7 5^{2} : 5625

25 x^{2} = 5625

Move

5625

to the left side

25 x^{2} - 5625 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 0 \pm 0 ^{2} - 4 \cdot 25 ( - 5625 )}{2 \cdot 25}

0^{2} - 4 \cdot 25 (- 5625) = 750

x_{1, 2} = \frac{- 0 \pm 750}{2 \cdot 25}

Separate the solutions

x_{1} = \frac{- 0 + 750}{2 \cdot 25}, x_{2} = \frac{- 0 - 750}{2 \cdot 25}

x = \frac{- 0 + 750}{2 \cdot 25} : 15

x = \frac{- 0 - 750}{2 \cdot 25} : - 15

The solutions to the quadratic equation are:

x = 15, x = - 15

2 t^{2} + 5 t + 2 = 0

y^{2} + 0 = 0

x, x^{2} + 15 x = - 50

2 x^{2} + 28 x + 80

x^{2} - 3 x + 6

What are the solutions to the equation (3x)^2+(4x)^2=75^2 ?
The solutions to the equation (3x)^2+(4x)^2=75^2 are x=15,x=-15
Find the zeros of (3x)^2+(4x)^2=75^2
The zeros of (3x)^2+(4x)^2=75^2 are x=15,x=-15