What are the solutions to the equation y^2-6ay+(9a^2-4)=0 ?

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

Find the zeros of y^2-6ay+(9a^2-4)=0

The zeros of y^2-6ay+(9a^2-4)=0 are y=3a+2,y=3a-2

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y^{2} - 6 a y + (9 a^{2} - 4) = 0

y = 3 a + 2, y = 3 a - 2

Solution steps

y^{2} - 6 a y + (9 a^{2} - 4) = 0

Refine

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

Solve with the quadratic formula

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

(- 6 a)^{2} - 4 \cdot 1 \cdot (9 a^{2} - 4) = 4

y_{1, 2} = \frac{- ( - 6 a ) \pm 4}{2 \cdot 1}

Separate the solutions

y_{1} = \frac{- ( - 6 a ) + 4}{2 \cdot 1}, y_{2} = \frac{- ( - 6 a ) - 4}{2 \cdot 1}

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

y = \frac{- ( - 6 a ) - 4}{2 \cdot 1} : 3 a - 2

The solutions to the quadratic equation are:

y = 3 a + 2, y = 3 a - 2

(x - 7)^{2} = (3 x + 4) (x - 6) + 1

x, x^{2} - 81 = 0

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

4 y^{2} + 3 = - 8 y

4 x^{2} - 20 + 25 = 12 + 25

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