# What is x ? (5x-6)^1/2=x

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To solve this (5x-6)^(1/2) = x:

We solve this by squaring both sides. We get a quadratic equation. We can solve the quadratic equation by making one side zero and expresion on the other side will be factored and ech of the flinear factors will be equated to zero.

5x-6 = x^2.

0 = x^2 -5x+6.

So x^2 -5x+6 = 0.

x^2- 3x-2x+6 = 0.

x(x-3) -2(x-3) = 0.

(x-3)(x-2) = 0.

x-3 = 0 , or x-2 = 0.

x-3 = gives x= 3.

x-2 = 0 gives x = 2.

Sp x= 2 or x= 3 are the solutions.

Before solving the equation, we'll impose conditions of existence of the square root.

5x-6 >= 0

We'll subtract 6 both sides:

5x >= 6

We'll divide by 5:

x >=6/5

The interval of admissible solutions for the given equation is:

[6/5 , +infinite)

Now, we'll solve the equation:

sqrt (5x-6) = x

We'll square raise both sides:

5x - 6 = x^2

We'll move all terms to one side and we'll use the symmetric property:

x^2 - 5x + 6 = 0

We'll apply the quadratic formula:

x1 = [5+sqrt(25+24)]/2

x2 = [5-sqrt(25+24)]/2

x1 = (5+1)/2

**x1 = 3**

x2 = (5-1)/2

**x2 = 2**

**Since both values belong to the interval of admissible values, they are accepted as solutions of the given equation.**