Question

solve for \x\
$(x^{2}-5x + 5)=1$

Explanation:

Step1: Rewrite the equation

We have the equation \((x^{2}-5x + 5)^{x^{2}-11x + 30}=1\). To solve this, we consider the cases where a number raised to a power equals 1:

• Case 1: The base is 1, and the exponent can be any real number. So we set \(x^{2}-5x + 5 = 1\).
• Case 2: The base is - 1, and the exponent is an even integer. So we set \(x^{2}-5x + 5=-1\) and check if the exponent \(x^{2}-11x + 30\) is even.
• Case 3: The exponent is 0, and the base is not 0. So we set \(x^{2}-11x + 30 = 0\) and check if \(x^{2}-5x + 5

eq0\).

Step2: Solve Case 1: \(x^{2}-5x + 5 = 1\)

Subtract 1 from both sides: \(x^{2}-5x+4 = 0\)
Factor the quadratic: \((x - 1)(x - 4)=0\)
Using the zero - product property, we get \(x-1 = 0\) or \(x - 4=0\), so \(x = 1\) or \(x = 4\).

Step3: Solve Case 2: \(x^{2}-5x + 5=-1\)

Add 1 to both sides: \(x^{2}-5x + 6 = 0\)
Factor the quadratic: \((x - 2)(x - 3)=0\)
Using the zero - product property, we get \(x-2 = 0\) or \(x - 3=0\), so \(x = 2\) or \(x = 3\).
Now we check the exponent for each value:

• When \(x = 2\), the exponent \(x^{2}-11x + 30=4-22 + 30 = 12\) (which is even).
• When \(x = 3\), the exponent \(x^{2}-11x + 30=9-33 + 30 = 6\) (which is even).

Step4: Solve Case 3: \(x^{2}-11x + 30 = 0\)

Factor the quadratic: \((x - 5)(x - 6)=0\)
Using the zero - product property, we get \(x-5 = 0\) or \(x - 6=0\), so \(x = 5\) or \(x = 6\).
Now we check the base for each value:

• When \(x = 5\), the base \(x^{2}-5x + 5=25-25 + 5 = 5

eq0\).

• When \(x = 6\), the base \(x^{2}-5x + 5=36-30 + 5 = 11

eq0\).

Answer:

The solutions of the equation \((x^{2}-5x + 5)^{x^{2}-11x + 30}=1\) are \(x = 1\), \(x = 2\), \(x = 3\), \(x = 4\), \(x = 5\), \(x = 6\).