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Gebruik de discriminant om volgende vierkantsvergelijkingen op te lossen

  1. \(\frac{1}{2}x^2-\frac{3}{4}x+\frac{9}{32}=0\)
  2. \(\frac{1}{3}x^2+\frac{1}{3}x-44=0\)
  3. \((5x-4)(3x-5)-x(9x-10)=26\)
  4. \(\frac{1}{32}x^2-\frac{1}{4}x+\frac{1}{2}=0\)
  5. \(-(9-2x)=-x^2-(73-18x)\)
  6. \(-\frac{1}{3}x=-\frac{4}{3}x^2-\frac{1}{3}\)
  7. \(x(4x+50)=2(x-72)\)
  8. \(8x^2+\frac{25}{6}x+\frac{1}{2}=0\)
  9. \((-3x+5)(-4x-1)-x(11x-24)=-125\)
  10. \(-(4-27x)=-18x^2-(6-14x)\)
  11. \((-3x-2)(-x+1)-x(x-10)=16\)
  12. \(\frac{1}{3}x=-\frac{1}{3}x^2+2\)

Gebruik de discriminant om volgende vierkantsvergelijkingen op te lossen

Verbetersleutel

  1. \(\frac{1}{2}x^2-\frac{3}{4}x+\frac{9}{32}=0\\ \Leftrightarrow \color{red}{32.} \left(\frac{1}{2}x^2-\frac{3}{4}x+\frac{9}{32}\right)=0 \color{red}{.32} \\ \Leftrightarrow 16x^2-24x+9=0 \\\text{We zoeken de oplossingen van } \color{blue}{16x^2-24x+9=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (-24)^2-4.16.9 & &\\ & = 576-576 & & \\ & = 0 & & \\ x & = \frac{-b\pm \sqrt{D}}{2.a} & & \\ & = \frac{-(-24)}{2.16} & & \\ & = \frac{3}{4} & & \\V &= \Big\{ \frac{3}{4} \Big\} & &\end{align} \\ -----------------\)
  2. \(\frac{1}{3}x^2+\frac{1}{3}x-44=0\\ \Leftrightarrow \color{red}{3.} \left(\frac{1}{3}x^2+\frac{1}{3}x-44\right)=0 \color{red}{.3} \\ \Leftrightarrow x^2+x-132=0 \\\text{We zoeken de oplossingen van } \color{blue}{x^2+x-132=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (1)^2-4.1.(-132) & &\\ & = 1+528 & & \\ & = 529 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-1-\sqrt529}{2.1} & & = \frac{-1+\sqrt529}{2.1} \\ & = \frac{-24}{2} & & = \frac{22}{2} \\ & = -12 & & = 11 \\ \\ V &= \Big\{ -12 ; 11 \Big\} & &\end{align} \\ -----------------\)
  3. \((5x-4)(3x-5)-x(9x-10)=26\\ \Leftrightarrow 15x^2-25x-12x+20 -9x^2+10x-26=0 \\ \Leftrightarrow 6x^2+5x-6=0 \\\text{We zoeken de oplossingen van } \color{blue}{6x^2+5x-6=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (5)^2-4.6.(-6) & &\\ & = 25+144 & & \\ & = 169 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-5-\sqrt169}{2.6} & & = \frac{-5+\sqrt169}{2.6} \\ & = \frac{-18}{12} & & = \frac{8}{12} \\ & = \frac{-3}{2} & & = \frac{2}{3} \\ \\ V &= \Big\{ \frac{-3}{2} ; \frac{2}{3} \Big\} & &\end{align} \\ -----------------\)
  4. \(\frac{1}{32}x^2-\frac{1}{4}x+\frac{1}{2}=0\\ \Leftrightarrow \color{red}{32.} \left(\frac{1}{32}x^2-\frac{1}{4}x+\frac{1}{2}\right)=0 \color{red}{.32} \\ \Leftrightarrow 9x^2-72x+144=0 \\\text{We zoeken de oplossingen van } \color{blue}{9x^2-72x+144=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (-72)^2-4.9.144 & &\\ & = 5184-5184 & & \\ & = 0 & & \\ x & = \frac{-b\pm \sqrt{D}}{2.a} & & \\ & = \frac{-(-72)}{2.9} & & \\ & = 4 & & \\V &= \Big\{ 4 \Big\} & &\end{align} \\ -----------------\)
  5. \(-(9-2x)=-x^2-(73-18x) \\ \Leftrightarrow -9+2x=-x^2-73+18x \\ \Leftrightarrow x^2-16x+64=0 \\\text{We zoeken de oplossingen van } \color{blue}{x^2-16x+64=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (-16)^2-4.1.64 & &\\ & = 256-256 & & \\ & = 0 & & \\ x & = \frac{-b\pm \sqrt{D}}{2.a} & & \\ & = \frac{-(-16)}{2.1} & & \\ & = 8 & & \\V &= \Big\{ 8 \Big\} & &\end{align} \\ -----------------\)
  6. \(-\frac{1}{3}x=-\frac{4}{3}x^2-\frac{1}{3} \\ \Leftrightarrow \frac{4}{3}x^2-\frac{1}{3}x+\frac{1}{3}=0 \\ \Leftrightarrow \color{red}{3.} \left(\frac{4}{3}x^2-\frac{1}{3}x+\frac{1}{3}\right)=0 \color{red}{.3} \\ \Leftrightarrow 16x^2-4x+4=0 \\\text{We zoeken de oplossingen van } \color{blue}{16x^2-4x+4=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (-4)^2-4.16.4 & &\\ & = 16-256 & & \\ & = -240 & & \\ & < 0 \\V &= \varnothing \end{align} \\ -----------------\)
  7. \(x(4x+50)=2(x-72) \\ \Leftrightarrow 4x^2+50x=2x-144 \\ \Leftrightarrow 4x^2+48x+144=0 \\\text{We zoeken de oplossingen van } \color{blue}{4x^2+48x+144=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (48)^2-4.4.144 & &\\ & = 2304-2304 & & \\ & = 0 & & \\ x & = \frac{-b\pm \sqrt{D}}{2.a} & & \\ & = \frac{-48}{2.4} & & \\ & = -6 & & \\V &= \Big\{ -6 \Big\} & &\end{align} \\ -----------------\)
  8. \(8x^2+\frac{25}{6}x+\frac{1}{2}=0\\ \Leftrightarrow \color{red}{6.} \left(8x^2+\frac{25}{6}x+\frac{1}{2}\right)=0 \color{red}{.6} \\ \Leftrightarrow 48x^2+25x+3=0 \\\text{We zoeken de oplossingen van } \color{blue}{48x^2+25x+3=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (25)^2-4.48.3 & &\\ & = 625-576 & & \\ & = 49 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-25-\sqrt49}{2.48} & & = \frac{-25+\sqrt49}{2.48} \\ & = \frac{-32}{96} & & = \frac{-18}{96} \\ & = \frac{-1}{3} & & = \frac{-3}{16} \\ \\ V &= \Big\{ \frac{-1}{3} ; \frac{-3}{16} \Big\} & &\end{align} \\ -----------------\)
  9. \((-3x+5)(-4x-1)-x(11x-24)=-125\\ \Leftrightarrow 12x^2+3x-20x-5 -11x^2+24x+125=0 \\ \Leftrightarrow x^2+22x+120=0 \\\text{We zoeken de oplossingen van } \color{blue}{x^2+22x+120=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (22)^2-4.1.120 & &\\ & = 484-480 & & \\ & = 4 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-22-\sqrt4}{2.1} & & = \frac{-22+\sqrt4}{2.1} \\ & = \frac{-24}{2} & & = \frac{-20}{2} \\ & = -12 & & = -10 \\ \\ V &= \Big\{ -12 ; -10 \Big\} & &\end{align} \\ -----------------\)
  10. \(-(4-27x)=-18x^2-(6-14x) \\ \Leftrightarrow -4+27x=-18x^2-6+14x \\ \Leftrightarrow 18x^2+13x+2=0 \\\text{We zoeken de oplossingen van } \color{blue}{18x^2+13x+2=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (13)^2-4.18.2 & &\\ & = 169-144 & & \\ & = 25 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-13-\sqrt25}{2.18} & & = \frac{-13+\sqrt25}{2.18} \\ & = \frac{-18}{36} & & = \frac{-8}{36} \\ & = \frac{-1}{2} & & = \frac{-2}{9} \\ \\ V &= \Big\{ \frac{-1}{2} ; \frac{-2}{9} \Big\} & &\end{align} \\ -----------------\)
  11. \((-3x-2)(-x+1)-x(x-10)=16\\ \Leftrightarrow 3x^2-3x+2x-2 -x^2+10x-16=0 \\ \Leftrightarrow 2x^2+5x-18=0 \\\text{We zoeken de oplossingen van } \color{blue}{2x^2+5x-18=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (5)^2-4.2.(-18) & &\\ & = 25+144 & & \\ & = 169 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-5-\sqrt169}{2.2} & & = \frac{-5+\sqrt169}{2.2} \\ & = \frac{-18}{4} & & = \frac{8}{4} \\ & = \frac{-9}{2} & & = 2 \\ \\ V &= \Big\{ \frac{-9}{2} ; 2 \Big\} & &\end{align} \\ -----------------\)
  12. \(\frac{1}{3}x=-\frac{1}{3}x^2+2 \\ \Leftrightarrow \frac{1}{3}x^2+\frac{1}{3}x-2=0 \\ \Leftrightarrow \color{red}{3.} \left(\frac{1}{3}x^2+\frac{1}{3}x-2\right)=0 \color{red}{.3} \\ \Leftrightarrow x^2+x-6=0 \\\text{We zoeken de oplossingen van } \color{blue}{x^2+x-6=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (1)^2-4.1.(-6) & &\\ & = 1+24 & & \\ & = 25 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-1-\sqrt25}{2.1} & & = \frac{-1+\sqrt25}{2.1} \\ & = \frac{-6}{2} & & = \frac{4}{2} \\ & = -3 & & = 2 \\ \\ V &= \Big\{ -3 ; 2 \Big\} & &\end{align} \\ -----------------\)
Oefeningengenerator vanhoeckes.be/wiskunde 2024-05-06 19:12:22