What are the solutions to the equation 9x^2-3x+a^2=0 ?

The solutions to the equation 9x^2-3x+a^2=0 are x=(1+sqrt(-4a^2+1))/6 ,x=(1-sqrt(-4a^2+1))/6

Find the zeros of 9x^2-3x+a^2=0

The zeros of 9x^2-3x+a^2=0 are x=(1+sqrt(-4a^2+1))/6 ,x=(1-sqrt(-4a^2+1))/6

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9 x^{2} - 3 x + a^{2} = 0

x = \frac{1 + - 4 a ^{2} + 1}{6}, x = \frac{1 - - 4 a ^{2} + 1}{6}

Solution steps

9 x^{2} - 3 x + a^{2} = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 9 a ^{2}}{2 \cdot 9}

Simplify

(- 3)^{2} - 4 \cdot 9 a^{2} : 3 1 - 4 a^{2}

x_{1, 2} = \frac{- ( - 3 ) \pm 3 1 - 4 a ^{2}}{2 \cdot 9}

Separate the solutions

x_{1} = \frac{- ( - 3 ) + 3 1 - 4 a ^{2}}{2 \cdot 9}, x_{2} = \frac{- ( - 3 ) - 3 1 - 4 a ^{2}}{2 \cdot 9}

x = \frac{- ( - 3 ) + 3 1 - 4 a ^{2}}{2 \cdot 9} : \frac{1 + - 4 a ^{2} + 1}{6}

x = \frac{- ( - 3 ) - 3 1 - 4 a ^{2}}{2 \cdot 9} : \frac{1 - - 4 a ^{2} + 1}{6}

The solutions to the quadratic equation are:

x = \frac{1 + - 4 a ^{2} + 1}{6}, x = \frac{1 - - 4 a ^{2} + 1}{6}

3 y^{2} + 11 y = 4

x (x + 8) = 884

x^{2} + (25) x + 5 = 0

5 y^{2} - y + 9 = 0

18 + x (x - 4) = 2 (x - 3)^{2}

What are the solutions to the equation 9x^2-3x+a^2=0 ?
The solutions to the equation 9x^2-3x+a^2=0 are x=(1+sqrt(-4a^2+1))/6 ,x=(1-sqrt(-4a^2+1))/6
Find the zeros of 9x^2-3x+a^2=0
The zeros of 9x^2-3x+a^2=0 are x=(1+sqrt(-4a^2+1))/6 ,x=(1-sqrt(-4a^2+1))/6