luk33
1

# Mam zadanie z matematyki rozwiąż równanie    16-(3x-1)do kwadratu=0   9xkwadrat-(2x-1)do kwadratu=0 (2x+1)do kwadratu=(4x+5)do kwadratu to pilne na jutro 10 punktów wiem tylko tyle że musi być tu delta i na koniec x1 i x2

(2) Odpowiedź
Glorianda

$a)\\ \\16-(3x-1)^2=0\\ \\16-(9x^2-6x+1)=0\\ \\16-9x^2+6x-1=0\\ \\-9x^2+6x+15=0\\ \\\Delta =b^2 -4ac =6^{2}-4*(-9)*15=36+540=576\\ \\\sqrt{\Delta }=\sqrt{576}=24$ $x_{1}=\frac{-b-\sqrt{\Delta }}{2a}=\frac{-6-24}{2*(-9)}=\frac{-30}{-18}=\frac{5}{3}=1\frac{2}{3}\\ \\x_{2}=\frac{-b+\sqrt{\Delta }}{2a}=\frac{-6+24}{2*(-9)}=\frac{18}{-18}= -1$ $b)\\ \\9x^2-(2x-1)^2=0\\ \\9x^2-(4x^2-4x+1)=0\\ \\9x^2-4x^2+4x-1=0\\ \\5x^2+4x-1=0\\ \\\Delta =b^2 -4ac =4^{2}-4*5*(-1)=16+20=36\\ \\\sqrt{\Delta }=\sqrt{36}=6$ $x_{1}=\frac{-b-\sqrt{\Delta }}{2a}=\frac{-4-6}{2*5}=\frac{-10}{10}=-1\\ \\x_{2}=\frac{-b+\sqrt{\Delta }}{2a}=\frac{-4+6}{2*5}=\frac{2}{10}=\frac{1}{5}$ $c)\\ \\(2x+1)^2=(4x+5)^2\\ \\4x^2+4x+1=16x^2+40x+25\\ \\4x^2+4x+1-16x^2-40x-25=0\\\\-12x^2-36x-24=0/:(-12)\\ \\x^{2}+3x+2=0\\ \\\Delta =b^{2}-4ac =3^2-4*1*2=9-8=1$ $\sqrt{\Delta }=\sqrt{1}=1\\ \\x_{1}=\frac{-b-\sqrt{\Delta }}{2a}=\frac{-3-1}{2}=\frac{-4}{2}=-2\\ \\x_{2}=\frac{-b+\sqrt{\Delta }}{2a}=\frac{-3+1}{2}=\frac{-2}{2}=-1$