What are the solutions to the equation solvefor x,9x^2+12xy+4y^2=0 ?

The solutions to the equation solvefor x,9x^2+12xy+4y^2=0 are x=-(2y)/3

Find the zeros of solvefor x,9x^2+12xy+4y^2=0

The zeros of solvefor x,9x^2+12xy+4y^2=0 are x=-(2y)/3

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solve for

x, 9 x^{2} + 12 x y + 4 y^{2} = 0

x = - \frac{2 y}{3}

Solution steps

9 x^{2} + 12 x y + 4 y^{2} = 0

Write in the standard form

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

9 x^{2} + 12 y x + 4 y^{2} = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 12 y \pm ( 12 y ) ^{2} - 4 \cdot 9 \cdot 4 y ^{2}}{2 \cdot 9}

(12 y)^{2} - 4 \cdot 9 \cdot 4 y^{2} = 0

x_{1, 2} = \frac{- 12 y \pm 0}{2 \cdot 9}

x = \frac{- 12 y}{2 \cdot 9}

Simplify

\frac{- 12 y}{2 \cdot 9} : - \frac{2 y}{3}

x = - \frac{2 y}{3}

The solution to the quadratic equation is:

x = - \frac{2 y}{3}

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

z^{2} + 3 z - 2 = 0

0 = 2 x - x^{2}

q, p = 2 + \frac{1}{5} q + \frac{1}{20} q^{2}

4 x^{2} - 3 x = 2

What are the solutions to the equation solvefor x,9x^2+12xy+4y^2=0 ?
The solutions to the equation solvefor x,9x^2+12xy+4y^2=0 are x=-(2y)/3
Find the zeros of solvefor x,9x^2+12xy+4y^2=0
The zeros of solvefor x,9x^2+12xy+4y^2=0 are x=-(2y)/3