If the inverse function of y = negative root minus x square is itself, then the definition domain of the original function is

If the inverse function of y = negative root minus x square is itself, then the definition domain of the original function is

-1

The inverse of x plus 6X equals Y is

X ^ 2 + 6x = y to get the inverse function of this, you have to try to deform the equation:
X^2+6X=X^2+6X=9-9=(X+3)^2-9
Therefore, the original formula can be changed into:
(X+3)^2-9=Y
(X+3)^2=Y+9
X + 3 = ± radical (y + 9)
X = ± radical (y + 9) - 3
I hope I can help you

Is the square of 2x minus 3x minus 2 equal to 0?

2x^2-3x-2=0
(2x+1)(x-2)=0
X = - 1 / 2 or x = 2

The square of 3x minus 1 equals 2x

3x^2-1=2x
3x^2-2x-1=0
(3x+1)(x-1)=0
X = - 1 / 3 or x = 1

2 x squared minus 3 x plus 1 equals 0

2x^2-3x+1=0
(2x-1)(x-1)=0
2x-1=0
x=1/2
x-1=0
X=1

Solution equation: the square of 2x minus 3x plus 1 equals 0

2x²-3x+1=0
Factorization factor of cross multiplication:
1 -1
X
2 -1
（x-1）（2x-1）=0
X-1 = 0 leads to x = 1
When 2x-1 = 0, x = 1 / 2
So:
x1=1
x2=1/2

2X squared plus 1 equals 3x

2x²+1=3x
2x²+1-3x=0
2x²-3x+1=0
(2x-1)(x-1)=0
2x-1 = 0 or X-1 = 0
X1 = 1 / 2 or x2 = 1

Find the algebraic value that the square of minus - 3x + 2x equals the square of - 3 -- 5x + 4x

The square of + 3x + 3x + 3 + is equal to (- 4x-3x)
(-3x+2x)²+(-3-5x+4x)²
=（-x）²+（-3-x）²
=2x²+6x+9

How to solve the problem that the square of 3x - 4x is equal to 2x?

3x^2-4x=2x
3x^2-6x=0
3x(x-2)=0
X = 0 or x = 2

If the square of 2x minus 6x + 1 equals 0, then what is the square of 6 + 3x-x thank you The square of 2x minus 6x + 1 is equal to 0, then 6 + 3x - (x square) is equal to several, urgent, urgent, urgent, urgent, urgent, urgent, urgent, urgent, urgent, urgent, urgent and urgent

2x^2-6x+1=0
2(x^2-3x)+1=0
x^2-3x=-1/2
6+3x-x^2=6+1/2=13/2