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Gebruik de discriminant om volgende vierkantsvergelijkingen op te lossen

  1. \(x(8x+15)=-2(x+1)\)
  2. \(x(x+37)=40(x+1)\)
  3. \((4x+4)(-2x+5)-x(-9x+32)=68\)
  4. \(3x^2+\frac{25}{6}x+\frac{4}{3}=0\)
  5. \(\frac{1}{2}x^2+\frac{13}{12}x+\frac{1}{2}=0\)
  6. \(2x^2-(18x-70)=x(x-1)\)
  7. \(\frac{1}{3}x^2-3x+\frac{14}{3}=0\)
  8. \(x(x+1)=8(x+1)\)
  9. \(17x^2-(15x-144)=x(x+81)\)
  10. \(\frac{1}{5}x^2+\frac{2}{5}x-\frac{63}{5}=0\)
  11. \(\frac{1}{8}x^2+\frac{1}{4}x+2=0\)
  12. \(\frac{1}{5}x=-\frac{1}{20}x^2+\frac{24}{5}\)

Gebruik de discriminant om volgende vierkantsvergelijkingen op te lossen

Verbetersleutel

  1. \(x(8x+15)=-2(x+1) \\ \Leftrightarrow 8x^2+15x=-2x-2 \\ \Leftrightarrow 8x^2+17x+2=0 \\\text{We zoeken de oplossingen van } \color{blue}{8x^2+17x+2=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (17)^2-4.8.2 & &\\ & = 289-64 & & \\ & = 225 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-17-\sqrt225}{2.8} & & = \frac{-17+\sqrt225}{2.8} \\ & = \frac{-32}{16} & & = \frac{-2}{16} \\ & = -2 & & = \frac{-1}{8} \\ \\ V &= \Big\{ -2 ; \frac{-1}{8} \Big\} & &\end{align} \\ -----------------\)
  2. \(x(x+37)=40(x+1) \\ \Leftrightarrow x^2+37x=40x+40 \\ \Leftrightarrow x^2-3x-40=0 \\\text{We zoeken de oplossingen van } \color{blue}{x^2-3x-40=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (-3)^2-4.1.(-40) & &\\ & = 9+160 & & \\ & = 169 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-(-3)-\sqrt169}{2.1} & & = \frac{-(-3)+\sqrt169}{2.1} \\ & = \frac{-10}{2} & & = \frac{16}{2} \\ & = -5 & & = 8 \\ \\ V &= \Big\{ -5 ; 8 \Big\} & &\end{align} \\ -----------------\)
  3. \((4x+4)(-2x+5)-x(-9x+32)=68\\ \Leftrightarrow -8x^2+20x-8x+20 +9x^2-32x-68=0 \\ \Leftrightarrow x^2+8x-48=0 \\\text{We zoeken de oplossingen van } \color{blue}{x^2+8x-48=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (8)^2-4.1.(-48) & &\\ & = 64+192 & & \\ & = 256 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-8-\sqrt256}{2.1} & & = \frac{-8+\sqrt256}{2.1} \\ & = \frac{-24}{2} & & = \frac{8}{2} \\ & = -12 & & = 4 \\ \\ V &= \Big\{ -12 ; 4 \Big\} & &\end{align} \\ -----------------\)
  4. \(3x^2+\frac{25}{6}x+\frac{4}{3}=0\\ \Leftrightarrow \color{red}{6.} \left(3x^2+\frac{25}{6}x+\frac{4}{3}\right)=0 \color{red}{.6} \\ \Leftrightarrow 18x^2+25x+8=0 \\\text{We zoeken de oplossingen van } \color{blue}{18x^2+25x+8=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (25)^2-4.18.8 & &\\ & = 625-576 & & \\ & = 49 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-25-\sqrt49}{2.18} & & = \frac{-25+\sqrt49}{2.18} \\ & = \frac{-32}{36} & & = \frac{-18}{36} \\ & = \frac{-8}{9} & & = \frac{-1}{2} \\ \\ V &= \Big\{ \frac{-8}{9} ; \frac{-1}{2} \Big\} & &\end{align} \\ -----------------\)
  5. \(\frac{1}{2}x^2+\frac{13}{12}x+\frac{1}{2}=0\\ \Leftrightarrow \color{red}{12.} \left(\frac{1}{2}x^2+\frac{13}{12}x+\frac{1}{2}\right)=0 \color{red}{.12} \\ \Leftrightarrow 6x^2+13x+6=0 \\\text{We zoeken de oplossingen van } \color{blue}{6x^2+13x+6=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (13)^2-4.6.6 & &\\ & = 169-144 & & \\ & = 25 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-13-\sqrt25}{2.6} & & = \frac{-13+\sqrt25}{2.6} \\ & = \frac{-18}{12} & & = \frac{-8}{12} \\ & = \frac{-3}{2} & & = \frac{-2}{3} \\ \\ V &= \Big\{ \frac{-3}{2} ; \frac{-2}{3} \Big\} & &\end{align} \\ -----------------\)
  6. \(2x^2-(18x-70)=x(x-1) \\ \Leftrightarrow 2x^2-18x+70=x^2-x \\ \Leftrightarrow x^2-17x+70=0 \\\text{We zoeken de oplossingen van } \color{blue}{x^2-17x+70=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (-17)^2-4.1.70 & &\\ & = 289-280 & & \\ & = 9 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-(-17)-\sqrt9}{2.1} & & = \frac{-(-17)+\sqrt9}{2.1} \\ & = \frac{14}{2} & & = \frac{20}{2} \\ & = 7 & & = 10 \\ \\ V &= \Big\{ 7 ; 10 \Big\} & &\end{align} \\ -----------------\)
  7. \(\frac{1}{3}x^2-3x+\frac{14}{3}=0\\ \Leftrightarrow \color{red}{3.} \left(\frac{1}{3}x^2-3x+\frac{14}{3}\right)=0 \color{red}{.3} \\ \Leftrightarrow x^2-9x+14=0 \\\text{We zoeken de oplossingen van } \color{blue}{x^2-9x+14=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (-9)^2-4.1.14 & &\\ & = 81-56 & & \\ & = 25 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-(-9)-\sqrt25}{2.1} & & = \frac{-(-9)+\sqrt25}{2.1} \\ & = \frac{4}{2} & & = \frac{14}{2} \\ & = 2 & & = 7 \\ \\ V &= \Big\{ 2 ; 7 \Big\} & &\end{align} \\ -----------------\)
  8. \(x(x+1)=8(x+1) \\ \Leftrightarrow x^2+x=8x+8 \\ \Leftrightarrow x^2-7x-8=0 \\\text{We zoeken de oplossingen van } \color{blue}{x^2-7x-8=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (-7)^2-4.1.(-8) & &\\ & = 49+32 & & \\ & = 81 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-(-7)-\sqrt81}{2.1} & & = \frac{-(-7)+\sqrt81}{2.1} \\ & = \frac{-2}{2} & & = \frac{16}{2} \\ & = -1 & & = 8 \\ \\ V &= \Big\{ -1 ; 8 \Big\} & &\end{align} \\ -----------------\)
  9. \(17x^2-(15x-144)=x(x+81) \\ \Leftrightarrow 17x^2-15x+144=x^2+81x \\ \Leftrightarrow 16x^2-96x+144=0 \\\text{We zoeken de oplossingen van } \color{blue}{16x^2-96x+144=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (-96)^2-4.16.144 & &\\ & = 9216-9216 & & \\ & = 0 & & \\ x & = \frac{-b\pm \sqrt{D}}{2.a} & & \\ & = \frac{-(-96)}{2.16} & & \\ & = 3 & & \\V &= \Big\{ 3 \Big\} & &\end{align} \\ -----------------\)
  10. \(\frac{1}{5}x^2+\frac{2}{5}x-\frac{63}{5}=0\\ \Leftrightarrow \color{red}{5.} \left(\frac{1}{5}x^2+\frac{2}{5}x-\frac{63}{5}\right)=0 \color{red}{.5} \\ \Leftrightarrow x^2+2x-63=0 \\\text{We zoeken de oplossingen van } \color{blue}{x^2+2x-63=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (2)^2-4.1.(-63) & &\\ & = 4+252 & & \\ & = 256 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-2-\sqrt256}{2.1} & & = \frac{-2+\sqrt256}{2.1} \\ & = \frac{-18}{2} & & = \frac{14}{2} \\ & = -9 & & = 7 \\ \\ V &= \Big\{ -9 ; 7 \Big\} & &\end{align} \\ -----------------\)
  11. \(\frac{1}{8}x^2+\frac{1}{4}x+2=0\\ \Leftrightarrow \color{red}{8.} \left(\frac{1}{8}x^2+\frac{1}{4}x+2\right)=0 \color{red}{.8} \\ \Leftrightarrow 4x^2+8x+64=0 \\\text{We zoeken de oplossingen van } \color{blue}{4x^2+8x+64=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (8)^2-4.4.64 & &\\ & = 64-1024 & & \\ & = -960 & & \\ & < 0 \\V &= \varnothing \end{align} \\ -----------------\)
  12. \(\frac{1}{5}x=-\frac{1}{20}x^2+\frac{24}{5} \\ \Leftrightarrow \frac{1}{20}x^2+\frac{1}{5}x-\frac{24}{5}=0 \\ \Leftrightarrow \color{red}{20.} \left(\frac{1}{20}x^2+\frac{1}{5}x-\frac{24}{5}\right)=0 \color{red}{.20} \\ \Leftrightarrow x^2+4x-96=0 \\\text{We zoeken de oplossingen van } \color{blue}{x^2+4x-96=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (4)^2-4.1.(-96) & &\\ & = 16+384 & & \\ & = 400 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-4-\sqrt400}{2.1} & & = \frac{-4+\sqrt400}{2.1} \\ & = \frac{-24}{2} & & = \frac{16}{2} \\ & = -12 & & = 8 \\ \\ V &= \Big\{ -12 ; 8 \Big\} & &\end{align} \\ -----------------\)
Oefeningengenerator wiskundeoefeningen.be 2025-12-05 21:43:44
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