What are the solutions to the equation 9x^2-6a^2x-b^4+a^4=0 ?

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

Find the zeros of 9x^2-6a^2x-b^4+a^4=0

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

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9 x^{2} - 6 a^{2} x - b^{4} + a^{4} = 0

x = \frac{a ^{2} + b ^{2}}{3}, x = \frac{a ^{2} - b ^{2}}{3}

Solution steps

9 x^{2} - 6 a^{2} x - b^{4} + a^{4} = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 6 a ^{2} ) \pm ( - 6 a ^{2} ) ^{2} - 4 \cdot 9 ( - b ^{4} + a ^{4} )}{2 \cdot 9}

Simplify

(- 6 a^{2})^{2} - 4 \cdot 9 (- b^{4} + a^{4}) : 6 b^{2}

x_{1, 2} = \frac{- ( - 6 a ^{2} ) \pm 6 b ^{2}}{2 \cdot 9}

Separate the solutions

x_{1} = \frac{- ( - 6 a ^{2} ) + 6 b ^{2}}{2 \cdot 9}, x_{2} = \frac{- ( - 6 a ^{2} ) - 6 b ^{2}}{2 \cdot 9}

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

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

The solutions to the quadratic equation are:

x = \frac{a ^{2} + b ^{2}}{3}, x = \frac{a ^{2} - b ^{2}}{3}

24 x^{2} + 18 x = 0

3 x^{2} + 15 x + 2 = 0

x, x^{2} + 2 y^{2} - 6 x - 8 y + 21 = 0

k^{2} - 8 k - 48 = 0

x^{2} + 2 x - 128 = 0

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