What are the solutions to the equation (4a)^2+(3a)^2=125^2 ?

The solutions to the equation (4a)^2+(3a)^2=125^2 are a=25,a=-25

Find the zeros of (4a)^2+(3a)^2=125^2

The zeros of (4a)^2+(3a)^2=125^2 are a=25,a=-25

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(4 a)^{2} + (3 a)^{2} = 12 5^{2}

a = 25, a = - 25

Solution steps

(4 a)^{2} + (3 a)^{2} = 12 5^{2}

Expand

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

Expand

12 5^{2} : 15625

25 a^{2} = 15625

Move

15625

to the left side

25 a^{2} - 15625 = 0

Solve with the quadratic formula

a_{1, 2} = \frac{- 0 \pm 0 ^{2} - 4 \cdot 25 ( - 15625 )}{2 \cdot 25}

0^{2} - 4 \cdot 25 (- 15625) = 1250

a_{1, 2} = \frac{- 0 \pm 1250}{2 \cdot 25}

Separate the solutions

a_{1} = \frac{- 0 + 1250}{2 \cdot 25}, a_{2} = \frac{- 0 - 1250}{2 \cdot 25}

a = \frac{- 0 + 1250}{2 \cdot 25} : 25

a = \frac{- 0 - 1250}{2 \cdot 25} : - 25

The solutions to the quadratic equation are:

a = 25, a = - 25

- 72 = - 18 x^{2}

(q) = 3 q^{2} + 5 q + 10

4.9 t^{2} + 12 t - 171 = 0

16 x^{2} - 102 + 44 = 0

z^{2} - 2 z = - 5

What are the solutions to the equation (4a)^2+(3a)^2=125^2 ?
The solutions to the equation (4a)^2+(3a)^2=125^2 are a=25,a=-25
Find the zeros of (4a)^2+(3a)^2=125^2
The zeros of (4a)^2+(3a)^2=125^2 are a=25,a=-25