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

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

Find the zeros of x^2-2ax=-(a^2-b^2)

The zeros of x^2-2ax=-(a^2-b^2) are x=a+b,x=a-b

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

x = a + b, x = a - b

Solution steps

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

Expand

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

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

Move

b^{2}

to the left side

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

Move

a^{2}

to the left side

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

Solve with the quadratic formula

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

Simplify

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

x = a + b, x = a - b

6 v^{2} - 30 = 0

x^{2} + 3 x (x - 2) = 5

3 x - 2 x^{2} = 7

t, y = \frac{1}{2} a t^{2}

(x + 2)^{2} = 11

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