What are the solutions to the equation λ^2-4λ+16(1)=0 ?

The solutions to the equation λ^2-4λ+16(1)=0 are λ=2+2sqrt(3)i,λ=2-2sqrt(3)i

Find the zeros of λ^2-4λ+16(1)=0

The zeros of λ^2-4λ+16(1)=0 are λ=2+2sqrt(3)i,λ=2-2sqrt(3)i

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

λ^{2} - 4 λ + 16 (1) = 0

λ = 2 + 23 i, λ = 2 - 23 i

Solution steps

λ^{2} - 4 λ + 16 (1) = 0

Expand

λ^{2} - 4 λ + 16 (1) : λ^{2} - 4 λ + 16

λ^{2} - 4 λ + 16 = 0

Solve with the quadratic formula

λ_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 1 \cdot 16}{2 \cdot 1}

Simplify

(- 4)^{2} - 4 \cdot 1 \cdot 16 : 43 i

λ_{1, 2} = \frac{- ( - 4 ) \pm 4 3 i}{2 \cdot 1}

Separate the solutions

λ_{1} = \frac{- ( - 4 ) + 4 3 i}{2 \cdot 1}, λ_{2} = \frac{- ( - 4 ) - 4 3 i}{2 \cdot 1}

λ = \frac{- ( - 4 ) + 4 3 i}{2 \cdot 1} : 2 + 23 i

λ = \frac{- ( - 4 ) - 4 3 i}{2 \cdot 1} : 2 - 23 i

The solutions to the quadratic equation are:

λ = 2 + 23 i, λ = 2 - 23 i

x^{2} + x - 110 = 0

(4 x - 1) (x - 3) = - 7

x, 36 x^{2} - 4 y^{2} = - 9 z^{2} + 216 x + 16 y - 272

x^{2} - 8 x - 1 = 8

3 x^{2} - 5 x - 1 = 0

What are the solutions to the equation λ^2-4λ+16(1)=0 ?
The solutions to the equation λ^2-4λ+16(1)=0 are λ=2+2sqrt(3)i,λ=2-2sqrt(3)i
Find the zeros of λ^2-4λ+16(1)=0
The zeros of λ^2-4λ+16(1)=0 are λ=2+2sqrt(3)i,λ=2-2sqrt(3)i