mariamikayla
Answered

Welcome to Westonci.ca, the Q&A platform where your questions are met with detailed answers from experienced experts. Join our platform to connect with experts ready to provide precise answers to your questions in various areas. Our platform provides a seamless experience for finding reliable answers from a network of experienced professionals.

Could you please solve me this exercice: :3

[tex]x^2-5x+\sqrt{x^2-5x+7}=5[/tex]

Thank you a lot! :3 :3 :3

Sagot :

[tex]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}[/tex]


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

[tex]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[/tex]

[tex]t+\sqrt{t}=12\ therefore\ t=16\ is\ not\ a\ solution.[/tex]


[tex]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=\frac{-b\pm\sqrt\Delta}{2a}\to x=\frac{5\pm\sqrt{33}}{2}[/tex]

[tex]Answer:\\\boxed{x=\frac{5-\sqrt{33}}{2}\ or\ x=\frac{5+\sqrt{33}}{2}}[/tex]
apologiabiology
 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
Thanks for stopping by. We strive to provide the best answers for all your questions. See you again soon. Thank you for your visit. We're dedicated to helping you find the information you need, whenever you need it. Westonci.ca is here to provide the answers you seek. Return often for more expert solutions.