What are the solutions to the equation x^2-(k+2)x+(k+5)=0 ?

The solutions to the equation x^2-(k+2)x+(k+5)=0 are x=(k+2+sqrt(k^2-16))/2 ,x=(k+2-sqrt(k^2-16))/2

Find the zeros of x^2-(k+2)x+(k+5)=0

The zeros of x^2-(k+2)x+(k+5)=0 are x=(k+2+sqrt(k^2-16))/2 ,x=(k+2-sqrt(k^2-16))/2

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x^{2} - (k + 2) x + (k + 5) = 0

x = \frac{k + 2 + k ^{2} - 16}{2}, x = \frac{k + 2 - k ^{2} - 16}{2}

Solution steps

x^{2} - (k + 2) x + (k + 5) = 0

Expand

x^{2} - (k + 2) x + (k + 5) : x^{2} - k x - 2 x + k + 5

x^{2} - k x - 2 x + k + 5 = 0

Write in the standard form

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

x^{2} - (k + 2) x + k + 5 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - k - 2 ) \pm ( - k - 2 ) ^{2} - 4 \cdot 1 \cdot ( k + 5 )}{2 \cdot 1}

Simplify

(- k - 2)^{2} - 4 \cdot 1 \cdot (k + 5) : k^{2} - 16

x_{1, 2} = \frac{- ( - k - 2 ) \pm k ^{2} - 16}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - k - 2 ) + k ^{2} - 16}{2 \cdot 1}, x_{2} = \frac{- ( - k - 2 ) - k ^{2} - 16}{2 \cdot 1}

x = \frac{- ( - k - 2 ) + k ^{2} - 16}{2 \cdot 1} : \frac{k + 2 + k ^{2} - 16}{2}

x = \frac{- ( - k - 2 ) - k ^{2} - 16}{2 \cdot 1} : \frac{k + 2 - k ^{2} - 16}{2}

The solutions to the quadratic equation are:

x = \frac{k + 2 + k ^{2} - 16}{2}, x = \frac{k + 2 - k ^{2} - 16}{2}

x, x^{2} (y - z) p + y^{2} (z - x) q = z^{2} (x - y)

x^{2} - 2 (a^{2} + b^{2}) x + (a^{2} - b^{2})^{2} = 0

a, a^{2} + b^{2} + 2 (3 a + 2 b) + 13 = 0

y, x = y^{2} + 7 z^{2}

p, z = p^{2} + q^{2}

What are the solutions to the equation x^2-(k+2)x+(k+5)=0 ?
The solutions to the equation x^2-(k+2)x+(k+5)=0 are x=(k+2+sqrt(k^2-16))/2 ,x=(k+2-sqrt(k^2-16))/2
Find the zeros of x^2-(k+2)x+(k+5)=0
The zeros of x^2-(k+2)x+(k+5)=0 are x=(k+2+sqrt(k^2-16))/2 ,x=(k+2-sqrt(k^2-16))/2