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

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

Find the zeros of 3x^2-2(a+b)x-(a+b)^2=0

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

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3 x^{2} - 2 (a + b) x - (a + b)^{2} = 0

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

Solution steps

3 x^{2} - 2 (a + b) x - (a + b)^{2} = 0

Expand

3 x^{2} - 2 (a + b) x - (a + b)^{2} : 3 x^{2} - 2 a x - 2 b x - a^{2} - 2 ab - b^{2}

3 x^{2} - 2 a x - 2 b x - a^{2} - 2 ab - b^{2} = 0

Write in the standard form

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

3 x^{2} - (2 a + 2 b) x - a^{2} - 2 ab - b^{2} = 0

Solve with the quadratic formula

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

Simplify

(- 2 a - 2 b)^{2} - 4 \cdot 3 (- a^{2} - 2 ab - b^{2}) : 4 (a + b)

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

Separate the solutions

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

x = \frac{- ( - 2 a - 2 b ) + 4 ( a + b )}{2 \cdot 3} : a + b

x = \frac{- ( - 2 a - 2 b ) - 4 ( a + b )}{2 \cdot 3} : - \frac{a + b}{3}

The solutions to the quadratic equation are:

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

\frac{( y - 2 ) ^{2}}{3} = 8 - y

4 x^{2} + 4 x - 80 = 0

x, 2 x^{2} - 3 x - 5 = 0

x^{2} + 6 x + 107 = 0

\frac{70 - 2 x ^{2}}{a} = 5

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