What are the solutions to the equation 1-18p+81p^2=0 ?

The solutions to the equation 1-18p+81p^2=0 are p= 1/9

Find the zeros of 1-18p+81p^2=0

The zeros of 1-18p+81p^2=0 are p= 1/9

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1 - 18 p + 81 p^{2} = 0

p = \frac{1}{9}

Solution steps

1 - 18 p + 81 p^{2} = 0

Write in the standard form

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

81 p^{2} - 18 p + 1 = 0

Solve with the quadratic formula

p_{1, 2} = \frac{- ( - 18 ) \pm ( - 18 ) ^{2} - 4 \cdot 81 \cdot 1}{2 \cdot 81}

(- 18)^{2} - 4 \cdot 81 \cdot 1 = 0

p_{1, 2} = \frac{- ( - 18 ) \pm 0}{2 \cdot 81}

p = \frac{- ( - 18 )}{2 \cdot 81}

\frac{- ( - 18 )}{2 \cdot 81} = \frac{1}{9}

p = \frac{1}{9}

The solution to the quadratic equation is:

p = \frac{1}{9}

x^{4} - 2 x^{3} + x^{2} + 12 x + 8

y^{2} + x y - 2 x^{2}

- 2 (x + 6) \geq 3 x - x + 4

16 x - 14 \leq 66

\frac{5 x - 14}{- 3} = - 7

What are the solutions to the equation 1-18p+81p^2=0 ?
The solutions to the equation 1-18p+81p^2=0 are p= 1/9
Find the zeros of 1-18p+81p^2=0
The zeros of 1-18p+81p^2=0 are p= 1/9