VKV met breuken of rekenwerk

Hoofdmenu Eentje per keer 

Gebruik de discriminant om volgende vierkantsvergelijkingen op te lossen

  1. \(\frac{1}{12}x^2+\frac{1}{3}x-5=0\)
  2. \(\frac{1}{3}x^2+\frac{10}{3}x-\frac{11}{3}=0\)
  3. \(\frac{1}{12}x^2+\frac{1}{4}x-9=0\)
  4. \((-3x+5)(-3x+5)-x(5x+22)=16\)
  5. \(x(x-5)=-12(x+1)\)
  6. \(\frac{1}{4}x^2-\frac{15}{4}x+11=0\)
  7. \(\frac{1}{20}x^2+\frac{1}{5}x-\frac{3}{5}=0\)
  8. \(\frac{1}{2}x^2+\frac{17}{8}x+\frac{1}{2}=0\)
  9. \((4x-5)(-4x-1)-x(-24x-24)=-13\)
  10. \(\frac{2}{5}x^2+\frac{3}{5}x+\frac{5}{2}=0\)
  11. \(-(14+x)=-x^2-(114-19x)\)
  12. \(-(11-27x)=-2x^2-(29-14x)\)

Gebruik de discriminant om volgende vierkantsvergelijkingen op te lossen

Verbetersleutel

  1. \(\frac{1}{12}x^2+\frac{1}{3}x-5=0\\ \Leftrightarrow \color{red}{12.} \left(\frac{1}{12}x^2+\frac{1}{3}x-5\right)=0 \color{red}{.12} \\ \Leftrightarrow x^2+4x-60=0 \\\text{We zoeken de oplossingen van } \color{blue}{x^2+4x-60=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (4)^2-4.1.(-60) & &\\ & = 16+240 & & \\ & = 256 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-4-\sqrt256}{2.1} & & = \frac{-4+\sqrt256}{2.1} \\ & = \frac{-20}{2} & & = \frac{12}{2} \\ & = -10 & & = 6 \\ \\ V &= \Big\{ -10 ; 6 \Big\} & &\end{align} \\ -----------------\)
  2. \(\frac{1}{3}x^2+\frac{10}{3}x-\frac{11}{3}=0\\ \Leftrightarrow \color{red}{3.} \left(\frac{1}{3}x^2+\frac{10}{3}x-\frac{11}{3}\right)=0 \color{red}{.3} \\ \Leftrightarrow x^2+10x-11=0 \\\text{We zoeken de oplossingen van } \color{blue}{x^2+10x-11=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (10)^2-4.1.(-11) & &\\ & = 100+44 & & \\ & = 144 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-10-\sqrt144}{2.1} & & = \frac{-10+\sqrt144}{2.1} \\ & = \frac{-22}{2} & & = \frac{2}{2} \\ & = -11 & & = 1 \\ \\ V &= \Big\{ -11 ; 1 \Big\} & &\end{align} \\ -----------------\)
  3. \(\frac{1}{12}x^2+\frac{1}{4}x-9=0\\ \Leftrightarrow \color{red}{12.} \left(\frac{1}{12}x^2+\frac{1}{4}x-9\right)=0 \color{red}{.12} \\ \Leftrightarrow x^2+3x-108=0 \\\text{We zoeken de oplossingen van } \color{blue}{x^2+3x-108=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (3)^2-4.1.(-108) & &\\ & = 9+432 & & \\ & = 441 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-3-\sqrt441}{2.1} & & = \frac{-3+\sqrt441}{2.1} \\ & = \frac{-24}{2} & & = \frac{18}{2} \\ & = -12 & & = 9 \\ \\ V &= \Big\{ -12 ; 9 \Big\} & &\end{align} \\ -----------------\)
  4. \((-3x+5)(-3x+5)-x(5x+22)=16\\ \Leftrightarrow 9x^2-15x-15x+25 -5x^2-22x-16=0 \\ \Leftrightarrow 4x^2-12x+9=0 \\\text{We zoeken de oplossingen van } \color{blue}{4x^2-12x+9=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (-12)^2-4.4.9 & &\\ & = 144-144 & & \\ & = 0 & & \\ x & = \frac{-b\pm \sqrt{D}}{2.a} & & \\ & = \frac{-(-12)}{2.4} & & \\ & = \frac{3}{2} & & \\V &= \Big\{ \frac{3}{2} \Big\} & &\end{align} \\ -----------------\)
  5. \(x(x-5)=-12(x+1) \\ \Leftrightarrow x^2-5x=-12x-12 \\ \Leftrightarrow x^2+7x+12=0 \\\text{We zoeken de oplossingen van } \color{blue}{x^2+7x+12=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (7)^2-4.1.12 & &\\ & = 49-48 & & \\ & = 1 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-7-\sqrt1}{2.1} & & = \frac{-7+\sqrt1}{2.1} \\ & = \frac{-8}{2} & & = \frac{-6}{2} \\ & = -4 & & = -3 \\ \\ V &= \Big\{ -4 ; -3 \Big\} & &\end{align} \\ -----------------\)
  6. \(\frac{1}{4}x^2-\frac{15}{4}x+11=0\\ \Leftrightarrow \color{red}{4.} \left(\frac{1}{4}x^2-\frac{15}{4}x+11\right)=0 \color{red}{.4} \\ \Leftrightarrow x^2-15x+44=0 \\\text{We zoeken de oplossingen van } \color{blue}{x^2-15x+44=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (-15)^2-4.1.44 & &\\ & = 225-176 & & \\ & = 49 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-(-15)-\sqrt49}{2.1} & & = \frac{-(-15)+\sqrt49}{2.1} \\ & = \frac{8}{2} & & = \frac{22}{2} \\ & = 4 & & = 11 \\ \\ V &= \Big\{ 4 ; 11 \Big\} & &\end{align} \\ -----------------\)
  7. \(\frac{1}{20}x^2+\frac{1}{5}x-\frac{3}{5}=0\\ \Leftrightarrow \color{red}{20.} \left(\frac{1}{20}x^2+\frac{1}{5}x-\frac{3}{5}\right)=0 \color{red}{.20} \\ \Leftrightarrow x^2+4x-12=0 \\\text{We zoeken de oplossingen van } \color{blue}{x^2+4x-12=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (4)^2-4.1.(-12) & &\\ & = 16+48 & & \\ & = 64 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-4-\sqrt64}{2.1} & & = \frac{-4+\sqrt64}{2.1} \\ & = \frac{-12}{2} & & = \frac{4}{2} \\ & = -6 & & = 2 \\ \\ V &= \Big\{ -6 ; 2 \Big\} & &\end{align} \\ -----------------\)
  8. \(\frac{1}{2}x^2+\frac{17}{8}x+\frac{1}{2}=0\\ \Leftrightarrow \color{red}{8.} \left(\frac{1}{2}x^2+\frac{17}{8}x+\frac{1}{2}\right)=0 \color{red}{.8} \\ \Leftrightarrow 4x^2+17x+4=0 \\\text{We zoeken de oplossingen van } \color{blue}{4x^2+17x+4=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (17)^2-4.4.4 & &\\ & = 289-64 & & \\ & = 225 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-17-\sqrt225}{2.4} & & = \frac{-17+\sqrt225}{2.4} \\ & = \frac{-32}{8} & & = \frac{-2}{8} \\ & = -4 & & = \frac{-1}{4} \\ \\ V &= \Big\{ -4 ; \frac{-1}{4} \Big\} & &\end{align} \\ -----------------\)
  9. \((4x-5)(-4x-1)-x(-24x-24)=-13\\ \Leftrightarrow -16x^2-4x+20x+5 +24x^2+24x+13=0 \\ \Leftrightarrow 8x^2+25x+18=0 \\\text{We zoeken de oplossingen van } \color{blue}{8x^2+25x+18=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (25)^2-4.8.18 & &\\ & = 625-576 & & \\ & = 49 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-25-\sqrt49}{2.8} & & = \frac{-25+\sqrt49}{2.8} \\ & = \frac{-32}{16} & & = \frac{-18}{16} \\ & = -2 & & = \frac{-9}{8} \\ \\ V &= \Big\{ -2 ; \frac{-9}{8} \Big\} & &\end{align} \\ -----------------\)
  10. \(\frac{2}{5}x^2+\frac{3}{5}x+\frac{5}{2}=0\\ \Leftrightarrow \color{red}{10.} \left(\frac{2}{5}x^2+\frac{3}{5}x+\frac{5}{2}\right)=0 \color{red}{.10} \\ \Leftrightarrow 4x^2+6x+25=0 \\\text{We zoeken de oplossingen van } \color{blue}{4x^2+6x+25=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (6)^2-4.4.25 & &\\ & = 36-400 & & \\ & = -364 & & \\ & < 0 \\V &= \varnothing \end{align} \\ -----------------\)
  11. \(-(14+x)=-x^2-(114-19x) \\ \Leftrightarrow -14-x=-x^2-114+19x \\ \Leftrightarrow x^2-20x+100=0 \\\text{We zoeken de oplossingen van } \color{blue}{x^2-20x+100=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (-20)^2-4.1.100 & &\\ & = 400-400 & & \\ & = 0 & & \\ x & = \frac{-b\pm \sqrt{D}}{2.a} & & \\ & = \frac{-(-20)}{2.1} & & \\ & = 10 & & \\V &= \Big\{ 10 \Big\} & &\end{align} \\ -----------------\)
  12. \(-(11-27x)=-2x^2-(29-14x) \\ \Leftrightarrow -11+27x=-2x^2-29+14x \\ \Leftrightarrow 2x^2+13x+18=0 \\\text{We zoeken de oplossingen van } \color{blue}{2x^2+13x+18=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (13)^2-4.2.18 & &\\ & = 169-144 & & \\ & = 25 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-13-\sqrt25}{2.2} & & = \frac{-13+\sqrt25}{2.2} \\ & = \frac{-18}{4} & & = \frac{-8}{4} \\ & = \frac{-9}{2} & & = -2 \\ \\ V &= \Big\{ \frac{-9}{2} ; -2 \Big\} & &\end{align} \\ -----------------\)
Oefeningengenerator wiskundeoefeningen.be 2026-11-10 14:14:41
Een site van Busleyden Atheneum Mechelen