What are the solutions to the equation ρ^2-3ρ+2=0 ?

The solutions to the equation ρ^2-3ρ+2=0 are ρ=2,ρ=1

Find the zeros of ρ^2-3ρ+2=0

The zeros of ρ^2-3ρ+2=0 are ρ=2,ρ=1

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ρ^{2} - 3 ρ + 2 = 0

ρ = 2, ρ = 1

Solution steps

ρ^{2} - 3 ρ + 2 = 0

Solve with the quadratic formula

ρ_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 1 \cdot 2}{2 \cdot 1}

(- 3)^{2} - 4 \cdot 1 \cdot 2 = 1

ρ_{1, 2} = \frac{- ( - 3 ) \pm 1}{2 \cdot 1}

Separate the solutions

ρ_{1} = \frac{- ( - 3 ) + 1}{2 \cdot 1}, ρ_{2} = \frac{- ( - 3 ) - 1}{2 \cdot 1}

ρ = \frac{- ( - 3 ) + 1}{2 \cdot 1} : 2

ρ = \frac{- ( - 3 ) - 1}{2 \cdot 1} : 1

The solutions to the quadratic equation are:

ρ = 2, ρ = 1

(x + 17)^{2} = - 4

16 = (x + 9)^{2}

6 x^{2} + 12 x - 12 = 0

10 a (1 + r + r^{2}) - 35 a r = 0

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

What are the solutions to the equation ρ^2-3ρ+2=0 ?
The solutions to the equation ρ^2-3ρ+2=0 are ρ=2,ρ=1
Find the zeros of ρ^2-3ρ+2=0
The zeros of ρ^2-3ρ+2=0 are ρ=2,ρ=1