What are the solutions to the equation (a-1)^2+a^2=5 ?

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

Find the zeros of (a-1)^2+a^2=5

The zeros of (a-1)^2+a^2=5 are a=2,a=-1

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(a - 1)^{2} + a^{2} = 5

a = 2, a = - 1

Solution steps

(a - 1)^{2} + a^{2} = 5

Expand

(a - 1)^{2} + a^{2} : 2 a^{2} - 2 a + 1

2 a^{2} - 2 a + 1 = 5

Move

5

to the left side

2 a^{2} - 2 a - 4 = 0

Solve with the quadratic formula

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

(- 2)^{2} - 4 \cdot 2 (- 4) = 6

a_{1, 2} = \frac{- ( - 2 ) \pm 6}{2 \cdot 2}

Separate the solutions

a_{1} = \frac{- ( - 2 ) + 6}{2 \cdot 2}, a_{2} = \frac{- ( - 2 ) - 6}{2 \cdot 2}

a = \frac{- ( - 2 ) + 6}{2 \cdot 2} : 2

a = \frac{- ( - 2 ) - 6}{2 \cdot 2} : - 1

The solutions to the quadratic equation are:

a = 2, a = - 1

(- 23)^{2}

- 12 x^{2} + 24 x

4 ∣ 2 z + 5 ∣ + 9 = 5

25 y = 15 y + 75

(y x)^{3}

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