What are the solutions to the equation a(x^2-15x+54)+b(x-9)=0,a+b ?

The solutions to the equation a(x^2-15x+54)+b(x-9)=0,a+b are x=9,x=(6a-b)/a ;\quad a\ne 0

Find the zeros of a(x^2-15x+54)+b(x-9)=0,a+b

The zeros of a(x^2-15x+54)+b(x-9)=0,a+b are x=9,x=(6a-b)/a ;\quad a\ne 0

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a (x^{2} - 15 x + 54) + b (x - 9) = 0, a + b

x = 9, x = \frac{6 a - b}{a}; a \neq = 0

Solution steps

a (x^{2} - 15 x + 54) + b (x - 9) = 0

Expand

a (x^{2} - 15 x + 54) + b (x - 9) : a x^{2} - 15 a x + 54 a + b x - 9 b

a x^{2} - 15 a x + 54 a + b x - 9 b = 0

Write in the standard form

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

a x^{2} + (- 15 a + b) x + 54 a - 9 b = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 15 a + b ) \pm ( - 15 a + b ) ^{2} - 4 a ( 54 a - 9 b )}{2 a}

Simplify

(- 15 a + b)^{2} - 4 a (54 a - 9 b) : 3 a + b

x_{1, 2} = \frac{- ( - 15 a + b ) \pm ( 3 a + b )}{2 a}; a \neq = 0

Separate the solutions

x_{1} = \frac{- ( - 15 a + b ) + 3 a + b}{2 a}, x_{2} = \frac{- ( - 15 a + b ) - ( 3 a + b )}{2 a}

x = \frac{- ( - 15 a + b ) + 3 a + b}{2 a} : 9

x = \frac{- ( - 15 a + b ) - ( 3 a + b )}{2 a} : \frac{6 a - b}{a}

The solutions to the quadratic equation are:

x = 9, x = \frac{6 a - b}{a}; a \neq = 0

0 = x^{2} + 8 x + 20

(2 y + 3) (y - 9) = 0

y, (1 + \frac{x}{2})^{2} + 3^{2} = y^{2}

- 4.9 x^{2} + 30 x + 200 = 0

(x^{2} - 2 x - 2) = 0

What are the solutions to the equation a(x^2-15x+54)+b(x-9)=0,a+b ?
The solutions to the equation a(x^2-15x+54)+b(x-9)=0,a+b are x=9,x=(6a-b)/a ;\quad a\ne 0
Find the zeros of a(x^2-15x+54)+b(x-9)=0,a+b
The zeros of a(x^2-15x+54)+b(x-9)=0,a+b are x=9,x=(6a-b)/a ;\quad a\ne 0