Question

Solve.


`√(x-1)+3=x`

Answer 1

`√(x-1)+3=x\\ √(x-1)=x-3\\ D:x-1\geq0 \wedge x-3\geq0\\ D:x\geq1 \wedge x\geq3\\ D:x\geq3\\ x-1=(x-3)^2\\ x-1=x^2-6x+9\\ x^2-7x+10=0\\ x^2-2x-5x+10=0\\ x(x-2)-5(x-2)=0\\ (x-5)(x-2)=0\\ x=5 \vee x=2\\ 2\\ot \in D \Rightarrow x=5`

Answer 2

`√(x-1)+3=x;\ Domain:x-1\geq0\ \wedge\ x\geq3\to x\geq1\ \wedge\ x\geq3\\D:x\in[3;\ \infty)\\\\√(x-1)=x-3\ \ \ \ |square\ both\ sides\\\\x-1=(x-3)^2\ \ \ \ |use\ of\ the\ formula:(a-b)^2=a^2-2ab+b^2\\\\x-1=x^2-6x+9\\\\x^2-6x+9=x-1\\\\x^2-6x-x+9+1=0\\x^2-7x+10=0\ \ \ \ |use\ quadratic\ formula`


`a=1;\ b=-7;\ c=10\\\\\Delta=b^2-4ac\to\Delta=(-7)^2-4\cdot1\cdot10=49-40=9\\\\x_1=(-b-\sqrt\Delta)/(2a);\ x_2=(-b+\sqrt\Delta)/(2a)\\\\\sqrt\Delta=\sqrt9=3\\\\x_1=(7-3)/(2\cdot1)=(4)/(2)=2\\otin D\\\\x_2=(7+3)/(2\cdot1)=(10)/(2)=5\in D\\\\Solution:x=5`