Question

$x^{2} -5x + \sqrt{ x^{2} -5x+7} =5$

Answer 1

$Domeniul: x^2-5x+7 \geq 0 \\\\ a=1 ; \ \ \ \ b=-5 ;\ \ \ \ c=7 \\\\ \Delta=b^2-4ac \\\\ \Delta=(-5)^2 -4*a*7 =25-28\to \boxed{-3} <0 \\\\ Dar \ D:x\in R \\\\ Aflam \ valoarea \ minima \ pentru \ :x^2-5x+7 \\\\ y_{min}= \frac{-\Delta}{4a}\to y_{min}=\frac{-(-3)}{4\cdot1}= \boxed{\frac{3}{4}} \\-------------------------------$

$x^2-5x+\sqrt{x^2-5x+7}=5 \ \ \ | (+7) Adaugam \ 7 \ in \ ambele \ parti\\\\ x^2-5x+7+\sqrt{x^2-5x+7}=12 \\\\ Inlocuim \ cu \ t=x^2-5x+7 \ \ \ (t \geq \frac{3}{4}) \\\\ t+\sqrt{t}=12 \ \ \ \ |(-t) \ \ \ Scadem \ t \ din \ ambele \ parti \\\\ \sqrt{t}=12-t \ \ \ |^2 \ \ \ \ Ridicam \ la \ patrat \ fiecare \ parte \\\\ t=(12-t)^2 \\\\ t=144-24t+t^2 \\\\ t^2-24t+144=t \ \ \ \ |(-t) \ \ \ \ Scadem \ t \ din \ fiecare \ parte \\\\ t^2-25t+144=0 \\\\ t^2-9t-16t+144=0 \\\\ t(t-9)-16(t-9)=0 \\\\ (t-9)(t-16)=0$

$t-9=0 \ \ \ or \ \ \ t-16=0 \\\\ t=9 \ \ \ \ or \ \ \ \ t=16 \\\\ t+\sqrt{t}=12 \ \ \ Rezulta \ ca \ t=16 \ nu \ poate \ fi \ solutie.\\\\\\ t=9 \to x^2-5x+7=9 \ \ \ \ |(-9) \ \ \ Scadem \ 9 \ din \ fiecare \ parte \\\\ x^2-5x-2=0 \\\\ a=1; \ \ \ \ b=-5; \ \ \ \ c=-2 \\\\ \Delta=(-5)^2-4\cdot1\cdot(-2)=25+8\to\boxed{33} \\\\ x= \frac{-b \pm \sqrt{\Delta}}{2a} \to x=\frac{5 \pm\sqrt{33}}{2} \\\\\\ Raspuns \ final: \\\\ \boxed{\boxed{x=\frac{5-\sqrt{33}}{2} \ \ \ or \ \ \ x=\frac{5+\sqrt{33}}{2}}}}$