What are the solutions to the equation x^2+ax-bx=ab ?

The solutions to the equation x^2+ax-bx=ab are x=b,x=-a

Find the zeros of x^2+ax-bx=ab

The zeros of x^2+ax-bx=ab are x=b,x=-a

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

x = b, x = - a

Solution steps

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

Move

ab

to the left side

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

Write in the standard form

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

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

Solve with the quadratic formula

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

Simplify

(a - b)^{2} - 4 \cdot 1 \cdot (- ab) : a + b

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

Separate the solutions

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

x = \frac{- ( a - b ) + a + b}{2 \cdot 1} : b

x = \frac{- ( a - b ) - ( a + b )}{2 \cdot 1} : - a

The solutions to the quadratic equation are:

x = b, x = - a

3 x^{2} - 60 x + 741 = 0

10 + x (x - 2) = 2

8 v^{2} + 8 = - 32

17 x^{2} + 719 x = 0

y, x = y^{2} - 1

What are the solutions to the equation x^2+ax-bx=ab ?
The solutions to the equation x^2+ax-bx=ab are x=b,x=-a
Find the zeros of x^2+ax-bx=ab
The zeros of x^2+ax-bx=ab are x=b,x=-a