What are the solutions to the equation x^2-mx+8=0,m,x_{1}=1 ?

The solutions to the equation x^2-mx+8=0,m,x_{1}=1 are x=(m+sqrt(m^2-32))/2 ,x=(m-sqrt(m^2-32))/2

Find the zeros of x^2-mx+8=0,m,x_{1}=1

The zeros of x^2-mx+8=0,m,x_{1}=1 are x=(m+sqrt(m^2-32))/2 ,x=(m-sqrt(m^2-32))/2

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x^{2} - m x + 8 = 0, m, x_{1} = 1

x = \frac{m + m ^{2} - 32}{2}, x = \frac{m - m ^{2} - 32}{2}

Solution steps

x^{2} - m x + 8 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - m ) \pm ( - m ) ^{2} - 4 \cdot 1 \cdot 8}{2 \cdot 1}

Simplify

(- m)^{2} - 4 \cdot 1 \cdot 8 : m^{2} - 32

x_{1, 2} = \frac{- ( - m ) \pm m ^{2} - 32}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - m ) + m ^{2} - 32}{2 \cdot 1}, x_{2} = \frac{- ( - m ) - m ^{2} - 32}{2 \cdot 1}

x = \frac{- ( - m ) + m ^{2} - 32}{2 \cdot 1} : \frac{m + m ^{2} - 32}{2}

x = \frac{- ( - m ) - m ^{2} - 32}{2 \cdot 1} : \frac{m - m ^{2} - 32}{2}

The solutions to the quadratic equation are:

x = \frac{m + m ^{2} - 32}{2}, x = \frac{m - m ^{2} - 32}{2}

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

x^{2} - 21 x + 38 = 0

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

- x^{2} - 1.8 \cdot 1 0^{- 5} x + 3.336 \cdot 1 0^{- 7} = 0

3 x^{2} + 39 x + 36 = 0

What are the solutions to the equation x^2-mx+8=0,m,x_{1}=1 ?
The solutions to the equation x^2-mx+8=0,m,x_{1}=1 are x=(m+sqrt(m^2-32))/2 ,x=(m-sqrt(m^2-32))/2
Find the zeros of x^2-mx+8=0,m,x_{1}=1
The zeros of x^2-mx+8=0,m,x_{1}=1 are x=(m+sqrt(m^2-32))/2 ,x=(m-sqrt(m^2-32))/2