Expression problem

Solve for $x$ :

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

#Algebra

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`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
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Comments

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 - 5 years, 3 months ago

Pretty sure you meant even in the third case.

Vishnu Bhagyanath - 5 years, 3 months ago

Yes, I wrote that incorrectly. Thanks.

Kay Xspre - 5 years, 3 months ago

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 - 5 years, 3 months ago

You mean: $(x^2 -7x + 12) x^2-3x +2 = 1$ $(x-3) (x-4) (x-1) (x-2)$ $x=3 x=4$ $x=1 x=2$ ?

A Former Brilliant Member - 5 years, 3 months ago

Saksham Srivastava - 5 years, 3 months ago

x^2-3x+2=0 => x^2-2x-x+2 =0 x= 2, x=1

Arnob Roy - 5 years, 3 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$