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Expression problem

Solve for \(x\):

\[ \large \displaystyle{(x^2-7x+12)^{x^2-3x+2} = 1} \]

Note by Cedie Camomot
6 months, 2 weeks ago

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Split into three cases

Case 1: \(x^2-7x+12 = 1\), and \(x^2-3x+2\) is any real number

The first case will give \(\displaystyle x = \frac{7\pm\sqrt{5}}{2}\), which, when substitute to \(x^2-3x+2\), gives the exponent of \(5\pm\sqrt{5}\), which is valid, as \(1^{5\pm\sqrt{5}} = 1\), hence one of the answers is \(\displaystyle x = \frac{7\pm\sqrt{5}}{2}\)


Case 2: \(x^2-7x+12\) is nonzero real number, and \(x^2-3x+2=0\)

This will give \(x = 1, 2\), and it is valid as \(6^0 = 2^0 = 1\).


Case 3: \(x^2-7x+12 = -1\), and \(x^2-3x+2\) is even numbers

This will give \(\displaystyle x = \frac{7\pm\sqrt{3}i}{2}\) (where \(i = \sqrt{-1}\)), and the exponents is \(3\pm\sqrt{3}i\). It will be \((-1)^{3\pm\sqrt{3}i}\), which gives a result in transcendental number and never equals 1


Therefore, there are four solutions, which is \(1, 2, \phi+3, 4-\phi\) (where \(\displaystyle \phi = \frac{1+\sqrt{5}}{2}\)) Kay Xspre · 6 months, 2 weeks ago

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@Kay Xspre Pretty sure you meant even in the third case. Vishnu Bhagyanath · 6 months, 2 weeks ago

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@Vishnu Bhagyanath Yes, I wrote that incorrectly. Thanks. Kay Xspre · 6 months, 2 weeks ago

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Use factoring algebra. Like this : (x^2 -7x + 12) x^2-3x +2 = 1 (x-3) (x-4) (x-1) (x-2) (x=3 x=4 ) (x=1 x=2) Try to apply one by one the value of x in the algebra problems Anggun Lestari · 6 months, 2 weeks ago

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@Anggun Lestari You mean: \( (x^2 -7x + 12) x^2-3x +2 = 1\) \((x-3) (x-4) (x-1) (x-2)\) \(x=3 x=4\) \(x=1 x=2 \)? Cedie Camomot · 6 months, 2 weeks ago

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1 Saksham Srivastava · 6 months, 1 week ago

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x^2-3x+2=0 => x^2-2x-x+2 =0 x= 2, x=1 Arnob Roy · 6 months, 2 weeks ago

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