What are the solutions to the equation a^2+(-2a)^2=40^2 ?

The solutions to the equation a^2+(-2a)^2=40^2 are a=8sqrt(5),a=-8sqrt(5)

Find the zeros of a^2+(-2a)^2=40^2

The zeros of a^2+(-2a)^2=40^2 are a=8sqrt(5),a=-8sqrt(5)

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a^{2} + (- 2 a)^{2} = 4 0^{2}

a = 85, a = - 85

Decimal Notation

a = 17.88854 \dots, a = - 17.88854 \dots

Solution steps

a^{2} + (- 2 a)^{2} = 4 0^{2}

Expand

a^{2} + (- 2 a)^{2} : 5 a^{2}

Expand

4 0^{2} : 1600

5 a^{2} = 1600

Move

1600

to the left side

5 a^{2} - 1600 = 0

Solve with the quadratic formula

a_{1, 2} = \frac{- 0 \pm 0 ^{2} - 4 \cdot 5 ( - 1600 )}{2 \cdot 5}

0^{2} - 4 \cdot 5 (- 1600) = 805

a_{1, 2} = \frac{- 0 \pm 80 5}{2 \cdot 5}

Separate the solutions

a_{1} = \frac{- 0 + 80 5}{2 \cdot 5}, a_{2} = \frac{- 0 - 80 5}{2 \cdot 5}

a = \frac{- 0 + 80 5}{2 \cdot 5} : 85

a = \frac{- 0 - 80 5}{2 \cdot 5} : - 85

The solutions to the quadratic equation are:

a = 85, a = - 85

2 x^{2} - 20 x + 41 = 0

a x^{2}

x^{2} + 4 x + 4

t, s = u t + \frac{1}{2} a t^{2}

n^{2} + 17 n + 22 = 7

What are the solutions to the equation a^2+(-2a)^2=40^2 ?
The solutions to the equation a^2+(-2a)^2=40^2 are a=8sqrt(5),a=-8sqrt(5)
Find the zeros of a^2+(-2a)^2=40^2
The zeros of a^2+(-2a)^2=40^2 are a=8sqrt(5),a=-8sqrt(5)