Question

numarul a=(2x^2 -5x+4)(2x^2 -5x+8)+4 reprezinta patratul perfect al numarului...

Answer 1

     
$(2 x^{2}-5x+4)(2 x^{2} -5x+8)+4= \\ 4 x^{4} -10 x^{3}+8 x^{2} -10 x^{3}+25 x^{2} -20x+16 x^{2}-40x +32+4= \\ =4 x^{4} -20 x^{3}+49 x^{2} -60x +36 = (a x^{2} +bx+c)^{2} \\ (a x^{2} +bx+c)(a x^{2} +bx+c)= \\ =a^{2}x^{4} +ab x^{3}+ac x^{2} +ab x^{3}+ b^{2} x^{2}+bcx +ac x^{2} +bcx+ c^{2}= \\ =a^{2}x^{4} +2abx^{3}+(2ac+ b^{2})x^{2}+2bcx + c^{2}$


$=>4 x^{4} -20 x^{3}+49 x^{2} -60x +36 = \\ =a^{2}x^{4} +2abx^{3}+(2ac+ b^{2})x^{2}+2bcx + c^{2} \\ Scriem \; ecuatiile: \\ a^{2}=4 \\ 2ab=-20 \\ 2ac+ b^{2} = 49 \\ 2bc=-60 \\ c^{2}=36 \\ Rezolvare: \\ a = \sqrt{4}=\boxed{2} \\ 2ab=-20 \;=>\;b= \frac{-20}{2a}=\frac{-20}{2*2}= \frac{-20}{4}=\boxed{-5} \\ 2ac+ b^{2} = 49 \;=>\;2*2*c + (-5)^{2}=49\;=>\;c= \frac{49-25}{4}=\boxed{6} \\ 2bc=-60\;=>\;2*(-5)*6 =60\;=>\;-10*6=-60 \;Corect ! \\ c^{2} =36 \;=>\;6^{2}=36\;\; Corect!$


$4 x^{4} -20 x^{3}+49 x^{2} -60x +36 = (3x^{2} -5x+6)^{2} \\ =>\;a =(3x^{2} -5x+6)^{2} \\ =>\;a \;\;este\;\; patratul\;\; numarului: \;\; \boxed{3x^{2} -5x+6}$