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
·
1 year 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
·
1 year 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
·
1 year ago

## Comments

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TopNewestSplit into three cases

Case 1:\(x^2-7x+12 = 1\), and \(x^2-3x+2\) is any real numberThe 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 numbersThis 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 · 1 year ago

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– Vishnu Bhagyanath · 1 year ago

Pretty sure you meant even in the third case.Log in to reply

– Kay Xspre · 1 year ago

Yes, I wrote that incorrectly. Thanks.Log in to reply

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 · 1 year ago

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– Cedie Camomot · 1 year ago

You mean: \( (x^2 -7x + 12) x^2-3x +2 = 1\) \((x-3) (x-4) (x-1) (x-2)\) \(x=3 x=4\) \(x=1 x=2 \)?Log in to reply

1 – Saksham Srivastava · 1 year ago

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x^2-3x+2=0 => x^2-2x-x+2 =0 x= 2, x=1 – Arnob Roy · 1 year ago

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