Zadanie 2 i 5 w załącznikach.... Question from @PoMoCyY123

November 2018 1 18 Report

Zadanie 2 i 5 w załącznikach.

zad.2

$2x^2 + 3 = - 7x \\2x^2 +7x + 3 = 0\\\Delta =b^2-4ac = 7^2-4\cdot 2\cdot 3=49-24 = 25\\\sqrt{\Delta }=\sqrt{25}=5\\x_{1}=\frac{-b-\sqrt{\Delta }}{2a}=\frac{-7-5}{2\cdot 2}=\frac{-12}{4}=-3\\x_{2}=\frac{-b+\sqrt{\Delta }}{2a}=\frac{-7+5}{2\cdot 2}=\frac{-2}{4}=-\frac{1}{2}$

$x+10 = 3x^2\\3x^2-x-10 =0\\\Delta =b^2-4ac = (-1)^2-4\cdot 3\cdot (-10)=1+120 = 121\\\sqrt{\Delta }=\sqrt{121}=11\\x_{1}=\frac{-b-\sqrt{\Delta }}{2a}=\frac{1-11}{2\cdot 3}=\frac{-10}{6}=-\frac{5}{3}=-2\frac{2}{3}\\x_{2}=\frac{-b+\sqrt{\Delta }}{2a}=\frac{1+11}{2\cdot 3}=\frac{12}{6}= 2$

$16x^2+24\sqrt{2}·x = -18\\16x^2+24\sqrt{2}·x +18=0\\\Delta =b^2-4ac = (24\sqrt{2})^2-4\cdot 16\cdot 18=576\cdot 2-1152=1152-1152=0\\x_{0}=\frac{-b}{2a}=\frac{-24\sqrt{2}}{2\cdot 16}=\frac{-24\sqrt{2}}{32}=-\frac{3\sqrt{2}}{4}$

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$5x^2 =2-2x\\5x^2 +2x-2=0\\\Delta =b^2-4ac = 2^2-4\cdot 5\cdot (-2)=4+40=44\\\sqrt{\Delta }=\sqrt{44}=\sqrt{4\cdot 11}=2\sqrt{11}\\x_{1}=\frac{-b-\sqrt{\Delta }}{2a}=\frac{-2-2\sqrt{11}}{2\cdot 5}=\frac{2(-1-\sqrt{11})}{10}=\frac{-1-\sqrt{19}}{5}\\x_{2}=\frac{-b+\sqrt{\Delta }}{2a}=\frac{-2+2\sqrt{11}}{2\cdot 5}=\frac{2(\sqrt{11}-1)}{10}=\frac{\sqrt{11}-1}{5}$

$11(x^2+5) = x\\11x^2+55= x\\11x^2-x+55 = 0\\\Delta =b^2-4ac =(-1)^2-4\cdot 11\cdot 55=1-2420 = -2419$

∆<0 nie ma pierwiastków

$x^2+x = 4x+7\\x^2+x - 4x-7=0\\x^2-3x-7=0\\\Delta =b^2-4ac =(-3)^2-4\cdot 1\cdot(-7)=9+28=37\\\sqrt{\Delta }=\sqrt{37}\\x_{1}=\frac{-b-\sqrt{\Delta }}{2a}=\frac{3-\sqrt{37}}{2\cdot 1}=\frac{3-\sqrt{37}}{2}\\x_{2}=\frac{-b+\sqrt{\Delta }}{2a}=\frac{3+\sqrt{37}}{2\cdot 1}=\frac{3+\sqrt{37}}{2}$

$3x^2+1 = 7x\\3x^2-7x+1=0\\\Delta =b^2-4ac =(-7)^2-4\cdot 3\cdot1=49-12=37\\\sqrt{\Delta }=\sqrt{37}\\x_{1}=\frac{-b-\sqrt{\Delta }}{2a}=\frac{7-\sqrt{37}}{2\cdot 1}=\frac{7-\sqrt{37}}{2}\\x_{2}=\frac{-b+\sqrt{\Delta }}{2a}=\frac{7+\sqrt{37}}{2\cdot 1}=\frac{7+\sqrt{37}}{2}$

zad.5

$(2x-2)(x-3)<(x-3)(x-4)\\2x^2-6x-2x+6<x^2-4x-3x+12\\2x^2-6x-2x+6-x^2+4x+3x-12<0\\x^2-x-6<0\\\Delta =b^2-4ac = 1^2-4\cdot 1\cdotz(-6)=1+24=25\\\sqrt{\Delta }=\sqrt{5}=5\\x_{1}=\frac{-b-\sqrt{\Delta }}{2a}=\frac{1-5}{2\cdot 1}=\frac{-4}{2}= -2\\x_{2}=\frac{-b+\sqrt{\Delta }}{2a}=\frac{1+5}{2\cdot 1}=\frac{6}{2}=3$

a >0 ramiona paraboli do góry

$x\epsilon (-2,3)$

$(3x-1)(x-3)\geq (x-3)(2x-1)\\3x^2-9x-x+3\geq 2x^2-x-6x+3\\3x^2-9x-x+3-2x^2+x+6x-3\\x^2-x\geq 0\\x(x-1)\geqslant 0\\x\geq 0\wedge x-1\geq 0\\x\geq 0\wedge x\geq 1\\x\epsilon< 0,1>$

$(2x-2)^2(3x-2)(x-3)\\4x^2 -8x+4\geq 3x^2-9x-2x+6=0\\4x^2 -8x+4-3x^2+9x+2x-6\geq 0\\x^2+3x-2\geq 0\\\Delta =b^2-4ac = 3^2-4\cdot 1\cdotz(-2)=9+8=17\\\sqrt{\Delta }=\sqrt{17}\\x_{1}=\frac{-b-\sqrt{\Delta }}{2a}=\frac{-3-\sqrt{17}}{2\cdot 1}=\frac{-3-\sqrt{17}}{2}\\x_{2}=\frac{-b+\sqrt{\Delta }}{2a}=\frac{-3+\sqrt{17}}{2\cdot 1}=\frac{-3+\sqrt{17}}{2}$