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Gebruik de discriminant om volgende vierkantsvergelijkingen op te lossen

  1. \(\frac{9}{4}x^2-\frac{19}{2}x+25=0\)
  2. \(2x=-\frac{1}{2}x^2-\frac{9}{2}\)
  3. \(37x^2-(20x+4)=x(x-27)\)
  4. \(-\frac{1}{4}x=-\frac{1}{16}x^2-\frac{1}{4}\)
  5. \(x(x-65)=-84(x+1)\)
  6. \(\frac{3}{2}x=-2x^2+\frac{1}{2}\)
  7. \(\frac{2}{3}x^2-\frac{1}{3}x+\frac{1}{24}=0\)
  8. \((4x-3)(5x+2)-x(17x-23)=-54\)
  9. \(x(9x-91)=-49(x+1)\)
  10. \(-(15-17x)=-16x^2-(14-2x)\)
  11. \(2x^2-(14x-81)=x(x+4)\)
  12. \((2x+1)(5x-3)-x(9x-29)=-103\)

Gebruik de discriminant om volgende vierkantsvergelijkingen op te lossen

Verbetersleutel

  1. \(\frac{9}{4}x^2-\frac{19}{2}x+25=0\\ \Leftrightarrow \color{red}{4.} \left(\frac{9}{4}x^2-\frac{19}{2}x+25\right)=0 \color{red}{.4} \\ \Leftrightarrow 9x^2-38x+100=0 \\\text{We zoeken de oplossingen van } \color{blue}{9x^2-38x+100=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (-38)^2-4.9.100 & &\\ & = 1444-3600 & & \\ & = -2156 & & \\ & < 0 \\V &= \varnothing \end{align} \\ -----------------\)
  2. \(2x=-\frac{1}{2}x^2-\frac{9}{2} \\ \Leftrightarrow \frac{1}{2}x^2+2x+\frac{9}{2}=0 \\ \Leftrightarrow \color{red}{2.} \left(\frac{1}{2}x^2+2x+\frac{9}{2}\right)=0 \color{red}{.2} \\ \Leftrightarrow 9x^2+36x+81=0 \\\text{We zoeken de oplossingen van } \color{blue}{9x^2+36x+81=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (36)^2-4.9.81 & &\\ & = 1296-2916 & & \\ & = -1620 & & \\ & < 0 \\V &= \varnothing \end{align} \\ -----------------\)
  3. \(37x^2-(20x+4)=x(x-27) \\ \Leftrightarrow 37x^2-20x-4=x^2-27x \\ \Leftrightarrow 36x^2+7x-4=0 \\\text{We zoeken de oplossingen van } \color{blue}{36x^2+7x-4=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (7)^2-4.36.(-4) & &\\ & = 49+576 & & \\ & = 625 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-7-\sqrt625}{2.36} & & = \frac{-7+\sqrt625}{2.36} \\ & = \frac{-32}{72} & & = \frac{18}{72} \\ & = \frac{-4}{9} & & = \frac{1}{4} \\ \\ V &= \Big\{ \frac{-4}{9} ; \frac{1}{4} \Big\} & &\end{align} \\ -----------------\)
  4. \(-\frac{1}{4}x=-\frac{1}{16}x^2-\frac{1}{4} \\ \Leftrightarrow \frac{1}{16}x^2-\frac{1}{4}x+\frac{1}{4}=0 \\ \Leftrightarrow \color{red}{16.} \left(\frac{1}{16}x^2-\frac{1}{4}x+\frac{1}{4}\right)=0 \color{red}{.16} \\ \Leftrightarrow 9x^2-36x+36=0 \\\text{We zoeken de oplossingen van } \color{blue}{9x^2-36x+36=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (-36)^2-4.9.36 & &\\ & = 1296-1296 & & \\ & = 0 & & \\ x & = \frac{-b\pm \sqrt{D}}{2.a} & & \\ & = \frac{-(-36)}{2.9} & & \\ & = 2 & & \\V &= \Big\{ 2 \Big\} & &\end{align} \\ -----------------\)
  5. \(x(x-65)=-84(x+1) \\ \Leftrightarrow x^2-65x=-84x-84 \\ \Leftrightarrow x^2+19x+84=0 \\\text{We zoeken de oplossingen van } \color{blue}{x^2+19x+84=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (19)^2-4.1.84 & &\\ & = 361-336 & & \\ & = 25 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-19-\sqrt25}{2.1} & & = \frac{-19+\sqrt25}{2.1} \\ & = \frac{-24}{2} & & = \frac{-14}{2} \\ & = -12 & & = -7 \\ \\ V &= \Big\{ -12 ; -7 \Big\} & &\end{align} \\ -----------------\)
  6. \(\frac{3}{2}x=-2x^2+\frac{1}{2} \\ \Leftrightarrow 2x^2+\frac{3}{2}x-\frac{1}{2}=0 \\ \Leftrightarrow \color{red}{2.} \left(2x^2+\frac{3}{2}x-\frac{1}{2}\right)=0 \color{red}{.2} \\ \Leftrightarrow 4x^2+3x-1=0 \\\text{We zoeken de oplossingen van } \color{blue}{4x^2+3x-1=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (3)^2-4.4.(-1) & &\\ & = 9+16 & & \\ & = 25 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-3-\sqrt25}{2.4} & & = \frac{-3+\sqrt25}{2.4} \\ & = \frac{-8}{8} & & = \frac{2}{8} \\ & = -1 & & = \frac{1}{4} \\ \\ V &= \Big\{ -1 ; \frac{1}{4} \Big\} & &\end{align} \\ -----------------\)
  7. \(\frac{2}{3}x^2-\frac{1}{3}x+\frac{1}{24}=0\\ \Leftrightarrow \color{red}{24.} \left(\frac{2}{3}x^2-\frac{1}{3}x+\frac{1}{24}\right)=0 \color{red}{.24} \\ \Leftrightarrow 16x^2-8x+1=0 \\\text{We zoeken de oplossingen van } \color{blue}{16x^2-8x+1=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (-8)^2-4.16.1 & &\\ & = 64-64 & & \\ & = 0 & & \\ x & = \frac{-b\pm \sqrt{D}}{2.a} & & \\ & = \frac{-(-8)}{2.16} & & \\ & = \frac{1}{4} & & \\V &= \Big\{ \frac{1}{4} \Big\} & &\end{align} \\ -----------------\)
  8. \((4x-3)(5x+2)-x(17x-23)=-54\\ \Leftrightarrow 20x^2+8x-15x-6 -17x^2+23x+54=0 \\ \Leftrightarrow 3x^2+25x+48=0 \\\text{We zoeken de oplossingen van } \color{blue}{3x^2+25x+48=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (25)^2-4.3.48 & &\\ & = 625-576 & & \\ & = 49 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-25-\sqrt49}{2.3} & & = \frac{-25+\sqrt49}{2.3} \\ & = \frac{-32}{6} & & = \frac{-18}{6} \\ & = \frac{-16}{3} & & = -3 \\ \\ V &= \Big\{ \frac{-16}{3} ; -3 \Big\} & &\end{align} \\ -----------------\)
  9. \(x(9x-91)=-49(x+1) \\ \Leftrightarrow 9x^2-91x=-49x-49 \\ \Leftrightarrow 9x^2-42x+49=0 \\\text{We zoeken de oplossingen van } \color{blue}{9x^2-42x+49=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (-42)^2-4.9.49 & &\\ & = 1764-1764 & & \\ & = 0 & & \\ x & = \frac{-b\pm \sqrt{D}}{2.a} & & \\ & = \frac{-(-42)}{2.9} & & \\ & = \frac{7}{3} & & \\V &= \Big\{ \frac{7}{3} \Big\} & &\end{align} \\ -----------------\)
  10. \(-(15-17x)=-16x^2-(14-2x) \\ \Leftrightarrow -15+17x=-16x^2-14+2x \\ \Leftrightarrow 16x^2+15x-1=0 \\\text{We zoeken de oplossingen van } \color{blue}{16x^2+15x-1=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (15)^2-4.16.(-1) & &\\ & = 225+64 & & \\ & = 289 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-15-\sqrt289}{2.16} & & = \frac{-15+\sqrt289}{2.16} \\ & = \frac{-32}{32} & & = \frac{2}{32} \\ & = -1 & & = \frac{1}{16} \\ \\ V &= \Big\{ -1 ; \frac{1}{16} \Big\} & &\end{align} \\ -----------------\)
  11. \(2x^2-(14x-81)=x(x+4) \\ \Leftrightarrow 2x^2-14x+81=x^2+4x \\ \Leftrightarrow x^2-18x+81=0 \\\text{We zoeken de oplossingen van } \color{blue}{x^2-18x+81=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (-18)^2-4.1.81 & &\\ & = 324-324 & & \\ & = 0 & & \\ x & = \frac{-b\pm \sqrt{D}}{2.a} & & \\ & = \frac{-(-18)}{2.1} & & \\ & = 9 & & \\V &= \Big\{ 9 \Big\} & &\end{align} \\ -----------------\)
  12. \((2x+1)(5x-3)-x(9x-29)=-103\\ \Leftrightarrow 10x^2-6x+5x-3 -9x^2+29x+103=0 \\ \Leftrightarrow x^2+20x+100=0 \\\text{We zoeken de oplossingen van } \color{blue}{x^2+20x+100=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (20)^2-4.1.100 & &\\ & = 400-400 & & \\ & = 0 & & \\ x & = \frac{-b\pm \sqrt{D}}{2.a} & & \\ & = \frac{-20}{2.1} & & \\ & = -10 & & \\V &= \Big\{ -10 \Big\} & &\end{align} \\ -----------------\)
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