What are the solutions to the equation (ax+5b)^2-(bx+5a)^2=0 ?

The solutions to the equation (ax+5b)^2-(bx+5a)^2=0 are x=-5,x=5;\quad a\ne-b,a\ne b

Find the zeros of (ax+5b)^2-(bx+5a)^2=0

The zeros of (ax+5b)^2-(bx+5a)^2=0 are x=-5,x=5;\quad a\ne-b,a\ne b

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(a x + 5 b)^{2} - (b x + 5 a)^{2} = 0

x = - 5, x = 5; a \neq = - b, a \neq = b

Solution steps

(a x + 5 b)^{2} - (b x + 5 a)^{2} = 0

Factor

(a x + 5 b)^{2} - (b x + 5 a)^{2} : (a x + 5 a + 5 b + b x) (a x - 5 a + 5 b - b x)

(a x + 5 a + 5 b + b x) (a x - 5 a + 5 b - b x) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

a x + 5 a + 5 b + b x = 0 or a x - 5 a + 5 b - b x = 0

Solve

a x + 5 a + 5 b + b x = 0 : x = - 5; a \neq = - b

Solve

a x - 5 a + 5 b - b x = 0 : x = 5; a \neq = b

The solutions to the quadratic equation are:

x = - 5, x = 5; a \neq = - b, a \neq = b

x^{2} - 203 x + 252 = 0

x^{2} + 3 x - 18 = 0

x^{2} + 12 x + 20

x^{2} + 12 x + 13

8 y + 3 y^{2} = 35

What are the solutions to the equation (ax+5b)^2-(bx+5a)^2=0 ?
The solutions to the equation (ax+5b)^2-(bx+5a)^2=0 are x=-5,x=5;\quad a\ne-b,a\ne b
Find the zeros of (ax+5b)^2-(bx+5a)^2=0
The zeros of (ax+5b)^2-(bx+5a)^2=0 are x=-5,x=5;\quad a\ne-b,a\ne b