Mariamikaylaid
Answered

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]
Apologiabiologyid
 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

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