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

The solutions to the equation 2ax^2-(2+a)bx+b^2=0 are x= b/2 ,x= b/a ;\quad a\ne 0

Find the zeros of 2ax^2-(2+a)bx+b^2=0

The zeros of 2ax^2-(2+a)bx+b^2=0 are x= b/2 ,x= b/a ;\quad a\ne 0

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

x = \frac{b}{2}, x = \frac{b}{a}; a \neq = 0

Solution steps

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

Expand

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

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

Write in the standard form

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

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

Solve with the quadratic formula

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

Simplify

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

x_{1, 2} = \frac{- ( - 2 b - ab ) \pm b ( a - 2 )}{2 \cdot 2 a}; a \neq = 0

Separate the solutions

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

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

x = \frac{- ( - 2 b - ab ) - b ( a - 2 )}{2 \cdot 2 a} : \frac{b}{a}

The solutions to the quadratic equation are:

x = \frac{b}{2}, x = \frac{b}{a}; a \neq = 0

x^{2} - 6 x = - 3 x + 28

x^{2} - 100 = 3 x^{2} + 96

s^{2} + 12 s + 936 = 0

(2 x - 1)^{2} + 11 = 4

4 v^{2} + 9 v - 2 = 4 v

What are the solutions to the equation 2ax^2-(2+a)bx+b^2=0 ?
The solutions to the equation 2ax^2-(2+a)bx+b^2=0 are x= b/2 ,x= b/a ;\quad a\ne 0
Find the zeros of 2ax^2-(2+a)bx+b^2=0
The zeros of 2ax^2-(2+a)bx+b^2=0 are x= b/2 ,x= b/a ;\quad a\ne 0