What is the answer to sqrt(2x^2-6x+2)=sqrt(x^2-3x) ?

The answer to sqrt(2x^2-6x+2)=sqrt(x^2-3x) is x=2,x=1

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sqrt(2x^2-6x+2)=sqrt(x^2-3x)

2 x^{2} - 6 x + 2 = x^{2} - 3 x

x = 2, x = 1

Solution steps

2 x^{2} - 6 x + 2 = x^{2} - 3 x

Square both sides:

2 x^{2} - 6 x + 2 = x^{2} - 3 x

2 x^{2} - 6 x + 2 = x^{2} - 3 x

Solve

2 x^{2} - 6 x + 2 = x^{2} - 3 x : x = 2, x = 1

x = 2, x = 1

Verify Solutions:

x = 2

True

, x = 1

True

The solutions are

x = 2, x = 1

2 x + 11 = 3 x + 7

(\frac{x}{5 + 5})^{\frac{3}{2}} = 125

w - 5 = - 8 w + 73

x^{2} + 2 2 x - 6 = 0

2 x - 1 + 2 = \frac{( 3 x )}{( x - 1 )}

What is the answer to sqrt(2x^2-6x+2)=sqrt(x^2-3x) ?
The answer to sqrt(2x^2-6x+2)=sqrt(x^2-3x) is x=2,x=1
Solve for x: sqrt(2x^2-6x+2)=sqrt(x^2-3x)
The solution is x=2,x=1