Question

Could you please solve me this exercice: :3


`x^2-5x+√(x^2-5x+7)=5`

Thank you a lot! :3 :3 :3

Answer 1

if xy=0 then x or/and y must be 0 so we try to get it to zero later

so first we get the square root on one side by itself
so subtract x^2-5 from both sides (-x^2+5 to both sides)

√(x^2-5x+7)=-x^2+5x+5
square both sides (or both sides to the ^2 power)
gets rid of the √
x^2-5x+7=(-x^2+5x+5)^2=x^4-10x^3+15x^2+50x+25
so subtract x^2 from both sides
-5x+7=x^4-10x^3+14x^2+50x+25
add 5x to both sides
7=x^4-10x^3+14x^2+55x+25
subtract 7 from both sides
0=x^4-10x^3+14x^2+55x+18
factor
(x^2-5x-9)(x^2-5x-2)=0
(x^2-5x-9)=0
(x^2-5x-2)=0
I'm not sure how to proceed but if you can find the factors of (x^2-5x-2) and (x^2-5x-9) and set them to zero, you can get the answers

Answer 2

`Domain:x^2-5x+7\geq0\\\\a=1;\ b=-5;\ c=7\\\\\Delta=b^2-4ac\to\Delta=(-5)^2-4\cdot1\cdot7=25-28=-3 < 0\\\\therefore\ D:x\in\mathbb{R}`



`Find\ minimum\ value\ of\ x^2-5x+7\\\\y_(min)=(-\Delta)/(4a)\to y_(min)=(-(-3))/(4\cdot1)=\boxed{(3)/(4)}\\-----------------------------\\x^2-5x+√(x^2-5x+7)=5\ \ \ \ \ |add\ 7\ to\ both\ sides\\\\x^2-5x+7+√(x^2-5x+7)=12\\\\subtitute\ t=x^2-5x+7\ (t\geq(3)/(4))\\\\t+√(t)=12\ \ \ \ |subtract\ t\ from\ both\ sides\\\\√(t)=12-t\ \ \ \ \ \ |square\ both\ sides`


`t=(12-t)^2\\\\t=12^2-2\cdot12\cdot t+t^2\\\\t=144-24t+t^2\\\\t^2-24t+144=t\ \ \ \ \ |subtract\ t\ from\ both\ sides\\\\t^2-25t+144=0\\\\t^2-9t-16t+144=0\\\\t(t-9)-16(t-9)=0\\\\(t-9)(t-16)=0\iff t-9=0\ or\ t-16=0\\\\t=9\ or\ t=16`


`t+√(t)=12\ therefore\ t=16\ is\ not\ a\ solution.`



`t=9\to x^2-5x+7=9\ \ \ \ |subtract\ 9\ from\ both\ sides\\\\x^2-5x-2=0\\\\a=1;\ b=-5;\ c=-2\\\\\Delta=(-5)^2-4\cdot1\cdot(-2)=25+8=33\\\\x=(-b\pm\sqrt\Delta)/(2a)\to x=(5\pm√(33))/(2)`


`Answer:\\\boxed{x=(5-√(33))/(2)\ or\ x=(5+√(33))/(2)}`