What are the solutions to the equation x^2+3k+1=(k+2)x ?

The solutions to the equation x^2+3k+1=(k+2)x are x=(k+2+sqrt(k^2-8k))/2 ,x=(k+2-sqrt(k^2-8k))/2

Find the zeros of x^2+3k+1=(k+2)x

The zeros of x^2+3k+1=(k+2)x are x=(k+2+sqrt(k^2-8k))/2 ,x=(k+2-sqrt(k^2-8k))/2

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x^{2} + 3 k + 1 = (k + 2) x

x = \frac{k + 2 + k ^{2} - 8 k}{2}, x = \frac{k + 2 - k ^{2} - 8 k}{2}

Solution steps

x^{2} + 3 k + 1 = (k + 2) x

Move

(k + 2) x

to the left side

x^{2} + 3 k + 1 - (k + 2) x = 0

Write in the standard form

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

x^{2} - (k + 2) x + 3 k + 1 = 0

Solve with the quadratic formula

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

Simplify

(- k - 2)^{2} - 4 \cdot 1 \cdot (3 k + 1) : k^{2} - 8 k

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

x = \frac{k + 2 + k ^{2} - 8 k}{2}, x = \frac{k + 2 - k ^{2} - 8 k}{2}

x, (x^{5})^{2} = 7 y y^{5}

4 x^{2} - 8 x = 12

4 w^{2} + 19 w - 5 = 0

x (x - 1) = - 1

4 (t - 1)^{2} - 9 (t - 1) = - 2

What are the solutions to the equation x^2+3k+1=(k+2)x ?
The solutions to the equation x^2+3k+1=(k+2)x are x=(k+2+sqrt(k^2-8k))/2 ,x=(k+2-sqrt(k^2-8k))/2
Find the zeros of x^2+3k+1=(k+2)x
The zeros of x^2+3k+1=(k+2)x are x=(k+2+sqrt(k^2-8k))/2 ,x=(k+2-sqrt(k^2-8k))/2