What are the solutions to the equation x^2-2(k-1)x+6k-11=0 ?

The solutions to the equation x^2-2(k-1)x+6k-11=0 are x=k-1+sqrt(k^2-8k+12),x=k-1-sqrt(k^2-8k+12)

Find the zeros of x^2-2(k-1)x+6k-11=0

The zeros of x^2-2(k-1)x+6k-11=0 are x=k-1+sqrt(k^2-8k+12),x=k-1-sqrt(k^2-8k+12)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

x^{2} - 2 (k - 1) x + 6 k - 11 = 0

x = k - 1 + k^{2} - 8 k + 12, x = k - 1 - k^{2} - 8 k + 12

Solution steps

x^{2} - 2 (k - 1) x + 6 k - 11 = 0

Expand

x^{2} - 2 (k - 1) x + 6 k - 11 : x^{2} - 2 k x + 2 x + 6 k - 11

x^{2} - 2 k x + 2 x + 6 k - 11 = 0

Write in the standard form

a x^{2} + b x + c = 0

x^{2} + (- 2 k + 2) x + 6 k - 11 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 2 k + 2 ) \pm ( - 2 k + 2 ) ^{2} - 4 \cdot 1 \cdot ( 6 k - 11 )}{2 \cdot 1}

Simplify

(- 2 k + 2)^{2} - 4 \cdot 1 \cdot (6 k - 11) : 2 k^{2} - 8 k + 12

x_{1, 2} = \frac{- ( - 2 k + 2 ) \pm 2 k ^{2} - 8 k + 12}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 2 k + 2 ) + 2 k ^{2} - 8 k + 12}{2 \cdot 1}, x_{2} = \frac{- ( - 2 k + 2 ) - 2 k ^{2} - 8 k + 12}{2 \cdot 1}

x = \frac{- ( - 2 k + 2 ) + 2 k ^{2} - 8 k + 12}{2 \cdot 1} : k - 1 + k^{2} - 8 k + 12

x = \frac{- ( - 2 k + 2 ) - 2 k ^{2} - 8 k + 12}{2 \cdot 1} : k - 1 - k^{2} - 8 k + 12

The solutions to the quadratic equation are:

x = k - 1 + k^{2} - 8 k + 12, x = k - 1 - k^{2} - 8 k + 12

a^{2} + 2 1^{2} = 2 9^{2}

6 x^{2} + 8 x - 4 = 0

π x^{2} = 500

a^{2} = \frac{1}{9}

2 x^{2} - 3 x - (6 - 4 x) = 0

What are the solutions to the equation x^2-2(k-1)x+6k-11=0 ?
The solutions to the equation x^2-2(k-1)x+6k-11=0 are x=k-1+sqrt(k^2-8k+12),x=k-1-sqrt(k^2-8k+12)
Find the zeros of x^2-2(k-1)x+6k-11=0
The zeros of x^2-2(k-1)x+6k-11=0 are x=k-1+sqrt(k^2-8k+12),x=k-1-sqrt(k^2-8k+12)