What are the solutions to the equation (x^2+x-6)^2-(x^2+2x+1)^2=0 ?

The solutions to the equation (x^2+x-6)^2-(x^2+2x+1)^2=0 are x=1,x=-5/2 ,x=-7

Find the zeros of (x^2+x-6)^2-(x^2+2x+1)^2=0

The zeros of (x^2+x-6)^2-(x^2+2x+1)^2=0 are x=1,x=-5/2 ,x=-7

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(x^{2} + x - 6)^{2} - (x^{2} + 2 x + 1)^{2} = 0

x = 1, x = - \frac{5}{2}, x = - 7

Decimal Notation

x = 1, x = - 2.5, x = - 7

Solution steps

(x^{2} + x - 6)^{2} - (x^{2} + 2 x + 1)^{2} = 0

Factor

(x^{2} + x - 6)^{2} - (x^{2} + 2 x + 1)^{2} : (x - 1) (2 x + 5) (- x - 7)

(x - 1) (2 x + 5) (- x - 7) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

x - 1 = 0 or 2 x + 5 = 0 or - x - 7 = 0

Solve

x - 1 = 0 : x = 1

Solve

2 x + 5 = 0 : x = - \frac{5}{2}

Solve

- x - 7 = 0 : x = - 7

The solutions are

x = 1, x = - \frac{5}{2}, x = - 7

x (x - 1)^{3} (x - 2)^{5} = 0

(x - 15)^{5} - (x - 15)^{3} = 0

x, x^{2021} + y = 4 y - 3

- 9 x^{4} + 16 x^{2} = 0

\frac{a ^{100}}{17} = 13

What are the solutions to the equation (x^2+x-6)^2-(x^2+2x+1)^2=0 ?
The solutions to the equation (x^2+x-6)^2-(x^2+2x+1)^2=0 are x=1,x=-5/2 ,x=-7
Find the zeros of (x^2+x-6)^2-(x^2+2x+1)^2=0
The zeros of (x^2+x-6)^2-(x^2+2x+1)^2=0 are x=1,x=-5/2 ,x=-7