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Gebruik de discriminant om volgende vierkantsvergelijkingen op te lossen

  1. \(\frac{5}{4}x=-\frac{9}{4}x^2+1\)
  2. \(-(13-14x)=-x^2-(113-18x)\)
  3. \(2x^2-(17x-10)=x(x-6)\)
  4. \(14x^2-(2x-144)=13x(x-2)\)
  5. \(-(12+15x)=-x^2-(92-3x)\)
  6. \((-2x-3)(-3x+1)-x(2x-20)=1\)
  7. \((2x+1)(5x+2)-x(9x+2)=-14\)
  8. \(\frac{1}{2}x^2+\frac{25}{24}x+\frac{1}{2}=0\)
  9. \(x(x+13)=15(x+1)\)
  10. \(2x^2+\frac{13}{6}x+\frac{1}{2}=0\)
  11. \(\frac{1}{3}x^2+\frac{7}{12}x-3=0\)
  12. \(-(5-21x)=-x^2-(-1-20x)\)

Gebruik de discriminant om volgende vierkantsvergelijkingen op te lossen

Verbetersleutel

  1. \(\frac{5}{4}x=-\frac{9}{4}x^2+1 \\ \Leftrightarrow \frac{9}{4}x^2+\frac{5}{4}x-1=0 \\ \Leftrightarrow \color{red}{4.} \left(\frac{9}{4}x^2+\frac{5}{4}x-1\right)=0 \color{red}{.4} \\ \Leftrightarrow 9x^2+5x-4=0 \\\text{We zoeken de oplossingen van } \color{blue}{9x^2+5x-4=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (5)^2-4.9.(-4) & &\\ & = 25+144 & & \\ & = 169 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-5-\sqrt169}{2.9} & & = \frac{-5+\sqrt169}{2.9} \\ & = \frac{-18}{18} & & = \frac{8}{18} \\ & = -1 & & = \frac{4}{9} \\ \\ V &= \Big\{ -1 ; \frac{4}{9} \Big\} & &\end{align} \\ -----------------\)
  2. \(-(13-14x)=-x^2-(113-18x) \\ \Leftrightarrow -13+14x=-x^2-113+18x \\ \Leftrightarrow x^2-4x+100=0 \\\text{We zoeken de oplossingen van } \color{blue}{x^2-4x+100=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (-4)^2-4.1.100 & &\\ & = 16-400 & & \\ & = -384 & & \\ & < 0 \\V &= \varnothing \end{align} \\ -----------------\)
  3. \(2x^2-(17x-10)=x(x-6) \\ \Leftrightarrow 2x^2-17x+10=x^2-6x \\ \Leftrightarrow x^2-11x+10=0 \\\text{We zoeken de oplossingen van } \color{blue}{x^2-11x+10=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (-11)^2-4.1.10 & &\\ & = 121-40 & & \\ & = 81 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-(-11)-\sqrt81}{2.1} & & = \frac{-(-11)+\sqrt81}{2.1} \\ & = \frac{2}{2} & & = \frac{20}{2} \\ & = 1 & & = 10 \\ \\ V &= \Big\{ 1 ; 10 \Big\} & &\end{align} \\ -----------------\)
  4. \(14x^2-(2x-144)=13x(x-2) \\ \Leftrightarrow 14x^2-2x+144=13x^2-26x \\ \Leftrightarrow x^2+24x+144=0 \\\text{We zoeken de oplossingen van } \color{blue}{x^2+24x+144=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (24)^2-4.1.144 & &\\ & = 576-576 & & \\ & = 0 & & \\ x & = \frac{-b\pm \sqrt{D}}{2.a} & & \\ & = \frac{-24}{2.1} & & \\ & = -12 & & \\V &= \Big\{ -12 \Big\} & &\end{align} \\ -----------------\)
  5. \(-(12+15x)=-x^2-(92-3x) \\ \Leftrightarrow -12-15x=-x^2-92+3x \\ \Leftrightarrow x^2-18x+80=0 \\\text{We zoeken de oplossingen van } \color{blue}{x^2-18x+80=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (-18)^2-4.1.80 & &\\ & = 324-320 & & \\ & = 4 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-(-18)-\sqrt4}{2.1} & & = \frac{-(-18)+\sqrt4}{2.1} \\ & = \frac{16}{2} & & = \frac{20}{2} \\ & = 8 & & = 10 \\ \\ V &= \Big\{ 8 ; 10 \Big\} & &\end{align} \\ -----------------\)
  6. \((-2x-3)(-3x+1)-x(2x-20)=1\\ \Leftrightarrow 6x^2-2x+9x-3 -2x^2+20x-1=0 \\ \Leftrightarrow 4x^2+15x-4=0 \\\text{We zoeken de oplossingen van } \color{blue}{4x^2+15x-4=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (15)^2-4.4.(-4) & &\\ & = 225+64 & & \\ & = 289 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-15-\sqrt289}{2.4} & & = \frac{-15+\sqrt289}{2.4} \\ & = \frac{-32}{8} & & = \frac{2}{8} \\ & = -4 & & = \frac{1}{4} \\ \\ V &= \Big\{ -4 ; \frac{1}{4} \Big\} & &\end{align} \\ -----------------\)
  7. \((2x+1)(5x+2)-x(9x+2)=-14\\ \Leftrightarrow 10x^2+4x+5x+2 -9x^2-2x+14=0 \\ \Leftrightarrow x^2+4x+16=0 \\\text{We zoeken de oplossingen van } \color{blue}{x^2+4x+16=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (4)^2-4.1.16 & &\\ & = 16-64 & & \\ & = -48 & & \\ & < 0 \\V &= \varnothing \end{align} \\ -----------------\)
  8. \(\frac{1}{2}x^2+\frac{25}{24}x+\frac{1}{2}=0\\ \Leftrightarrow \color{red}{24.} \left(\frac{1}{2}x^2+\frac{25}{24}x+\frac{1}{2}\right)=0 \color{red}{.24} \\ \Leftrightarrow 12x^2+25x+12=0 \\\text{We zoeken de oplossingen van } \color{blue}{12x^2+25x+12=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (25)^2-4.12.12 & &\\ & = 625-576 & & \\ & = 49 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-25-\sqrt49}{2.12} & & = \frac{-25+\sqrt49}{2.12} \\ & = \frac{-32}{24} & & = \frac{-18}{24} \\ & = \frac{-4}{3} & & = \frac{-3}{4} \\ \\ V &= \Big\{ \frac{-4}{3} ; \frac{-3}{4} \Big\} & &\end{align} \\ -----------------\)
  9. \(x(x+13)=15(x+1) \\ \Leftrightarrow x^2+13x=15x+15 \\ \Leftrightarrow x^2-2x-15=0 \\\text{We zoeken de oplossingen van } \color{blue}{x^2-2x-15=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (-2)^2-4.1.(-15) & &\\ & = 4+60 & & \\ & = 64 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-(-2)-\sqrt64}{2.1} & & = \frac{-(-2)+\sqrt64}{2.1} \\ & = \frac{-6}{2} & & = \frac{10}{2} \\ & = -3 & & = 5 \\ \\ V &= \Big\{ -3 ; 5 \Big\} & &\end{align} \\ -----------------\)
  10. \(2x^2+\frac{13}{6}x+\frac{1}{2}=0\\ \Leftrightarrow \color{red}{6.} \left(2x^2+\frac{13}{6}x+\frac{1}{2}\right)=0 \color{red}{.6} \\ \Leftrightarrow 12x^2+13x+3=0 \\\text{We zoeken de oplossingen van } \color{blue}{12x^2+13x+3=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (13)^2-4.12.3 & &\\ & = 169-144 & & \\ & = 25 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-13-\sqrt25}{2.12} & & = \frac{-13+\sqrt25}{2.12} \\ & = \frac{-18}{24} & & = \frac{-8}{24} \\ & = \frac{-3}{4} & & = \frac{-1}{3} \\ \\ V &= \Big\{ \frac{-3}{4} ; \frac{-1}{3} \Big\} & &\end{align} \\ -----------------\)
  11. \(\frac{1}{3}x^2+\frac{7}{12}x-3=0\\ \Leftrightarrow \color{red}{12.} \left(\frac{1}{3}x^2+\frac{7}{12}x-3\right)=0 \color{red}{.12} \\ \Leftrightarrow 4x^2+7x-36=0 \\\text{We zoeken de oplossingen van } \color{blue}{4x^2+7x-36=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (7)^2-4.4.(-36) & &\\ & = 49+576 & & \\ & = 625 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-7-\sqrt625}{2.4} & & = \frac{-7+\sqrt625}{2.4} \\ & = \frac{-32}{8} & & = \frac{18}{8} \\ & = -4 & & = \frac{9}{4} \\ \\ V &= \Big\{ -4 ; \frac{9}{4} \Big\} & &\end{align} \\ -----------------\)
  12. \(-(5-21x)=-x^2-(-1-20x) \\ \Leftrightarrow -5+21x=-x^2+1+20x \\ \Leftrightarrow x^2+x-6=0 \\\text{We zoeken de oplossingen van } \color{blue}{x^2+x-6=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (1)^2-4.1.(-6) & &\\ & = 1+24 & & \\ & = 25 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-1-\sqrt25}{2.1} & & = \frac{-1+\sqrt25}{2.1} \\ & = \frac{-6}{2} & & = \frac{4}{2} \\ & = -3 & & = 2 \\ \\ V &= \Big\{ -3 ; 2 \Big\} & &\end{align} \\ -----------------\)
Oefeningengenerator vanhoeckes.be/wiskunde 2024-03-28 17:42:55