Solve irrational equation pls

Answers

Answer 1

Answer: $\hbox{Domain:}\nx^2+x-2\geq0 \wedge x^2-4x+3\geq0 \wedge x^2-1\geq0\nx^2-x+2x-2\geq0 \wedge x^2-x-3x+3\geq0 \wedge x^2\geq1\nx(x-1)+2(x-1)\geq 0 \wedge x(x-1)-3(x-1)\geq0 \wedge (x\geq 1 \vee x\leq-1)\n(x+2)(x-1)\geq0 \wedge (x-3)(x-1)\geq0\wedge x\in(-\infty,-1\rangle\cup\langle1,\infty)\nx\in(-\infty,-2\rangle\cup\langle1,\infty) \wedge x\in(-\infty,1\rangle \cup\langle3,\infty) \wedge x\in(-\infty,-1\rangle\cup\langle1,\infty)\nx\in(-\infty,-2\rangle\cup\langle3,\infty)$

$√(x^2+x-2)+√(x^2-4x+3)=√(x^2-1)\nx^2-1=x^2+x-2+2√((x^2+x-2)(x^2-4x+3))+x^2-4x+3\n2√((x^2+x-2)(x^2-4x+3))=-x^2+3x-2\n√((x^2+x-2)(x^2-4x+3))=(-x^2+3x-2)/(2)\n(x^2+x-2)(x^2-4x+3)=\left((-x^2+3x-2)/(2)\right)^2\n(x+2)(x-1)(x-3)(x-1)=\left((-x^2+x+2x-2)/(2)\right)^2\n(x+2)(x-3)(x-1)^2=\left((-x(x-1)+2(x-1))/(2)\right)^2\n(x+2)(x-3)(x-1)^2=\left((-(x-2)(x-1))/(2)\right)^2\n(x+2)(x-3)(x-1)^2=((x-2)^2(x-1)^2)/(4)\n4(x+2)(x-3)(x-1)^2=(x-2)^2(x-1)^2\n$
$4(x+2)(x-3)(x-1)^2-(x-2)^2(x-1)^2=0\n(x-1)^2(4(x+2)(x-3)-(x-2)^2)=0\n(x-1)^2(4(x^2-3x+2x-6)-(x^2-4x+4))=0\n(x-1)^2(4x^2-4x-24-x^2+4x-4)=0\n(x-1)^2(3x^2-28)=0\nx-1=0 \vee 3x^2-28=0\nx=1 \vee 3x^2=28\nx=1 \vee x^2=(28)/(3)\nx=1 \vee x=\sqrt{(28)/(3)} \vee x=-\sqrt{(28)/(3)}\n$

There's one more condition I forgot about
$-(x-2)(x-1)\geq0\nx\in\langle1,2\rangle\n$

Finally
$x\in(-\infty,-2\rangle\cup\langle3,\infty) \wedge x\in\langle1,2\rangle \wedge x=\{1,\sqrt{(28)/(3)}, -\sqrt{(28)/(3)}\}\n\boxed{\boxed{x=1}}$

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