What are the solutions to the equation 9x^2+αx+1=0 ?

The solutions to the equation 9x^2+αx+1=0 are x=(-α+sqrt(α^2-36))/(18),x=(-α-sqrt(α^2-36))/(18)

Find the zeros of 9x^2+αx+1=0

The zeros of 9x^2+αx+1=0 are x=(-α+sqrt(α^2-36))/(18),x=(-α-sqrt(α^2-36))/(18)

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9 x^{2} + αx + 1 = 0

x = \frac{- α + α ^{2} - 36}{18}, x = \frac{- α - α ^{2} - 36}{18}

Solution steps

9 x^{2} + αx + 1 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- α \pm α ^{2} - 4 \cdot 9 \cdot 1}{2 \cdot 9}

Simplify

α^{2} - 4 \cdot 9 \cdot 1 : α^{2} - 36

x_{1, 2} = \frac{- α \pm α ^{2} - 36}{2 \cdot 9}

Separate the solutions

x_{1} = \frac{- α + α ^{2} - 36}{2 \cdot 9}, x_{2} = \frac{- α - α ^{2} - 36}{2 \cdot 9}

x = \frac{- α + α ^{2} - 36}{2 \cdot 9} : \frac{- α + α ^{2} - 36}{18}

x = \frac{- α - α ^{2} - 36}{2 \cdot 9} : \frac{- α - α ^{2} - 36}{18}

The solutions to the quadratic equation are:

x = \frac{- α + α ^{2} - 36}{18}, x = \frac{- α - α ^{2} - 36}{18}

u^{2} - 7 u + 6 = 0

(x - 1)^{2} = x + 5

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

227 - 0.25 x^{2} = 2 + 2 x^{2}

32 x - 4 = 4 x^{2} + 60

What are the solutions to the equation 9x^2+αx+1=0 ?
The solutions to the equation 9x^2+αx+1=0 are x=(-α+sqrt(α^2-36))/(18),x=(-α-sqrt(α^2-36))/(18)
Find the zeros of 9x^2+αx+1=0
The zeros of 9x^2+αx+1=0 are x=(-α+sqrt(α^2-36))/(18),x=(-α-sqrt(α^2-36))/(18)