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Gebruik de discriminant om volgende vierkantsvergelijkingen op te lossen

  1. \(\frac{4}{25}x^2-\frac{4}{5}x+1=0\)
  2. \(10x^2-(16x-4)=x(x-29)\)
  3. \(\frac{4}{3}x^2+\frac{13}{9}x+\frac{1}{3}=0\)
  4. \(x(9x+23)=3(x-12)\)
  5. \(-\frac{23}{4}x=-\frac{1}{4}x^2-33\)
  6. \(18x^2+\frac{25}{4}x+\frac{1}{2}=0\)
  7. \((-5x+5)(2x+1)-x(-14x+38)=-95\)
  8. \(17x^2-(14x-4)=x(x-6)\)
  9. \((-5x+4)(-4x+5)-x(2x-10)=22\)
  10. \(3x^2-(20x-36)=2x(x-12)\)
  11. \(-(6-41x)=-18x^2-(14-16x)\)
  12. \(-(11-8x)=-12x^2-(8-3x)\)

Gebruik de discriminant om volgende vierkantsvergelijkingen op te lossen

Verbetersleutel

  1. \(\frac{4}{25}x^2-\frac{4}{5}x+1=0\\ \Leftrightarrow \color{red}{25.} \left(\frac{4}{25}x^2-\frac{4}{5}x+1\right)=0 \color{red}{.25} \\ \Leftrightarrow 4x^2-20x+25=0 \\\text{We zoeken de oplossingen van } \color{blue}{4x^2-20x+25=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (-20)^2-4.4.25 & &\\ & = 400-400 & & \\ & = 0 & & \\ x & = \frac{-b\pm \sqrt{D}}{2.a} & & \\ & = \frac{-(-20)}{2.4} & & \\ & = \frac{5}{2} & & \\V &= \Big\{ \frac{5}{2} \Big\} & &\end{align} \\ -----------------\)
  2. \(10x^2-(16x-4)=x(x-29) \\ \Leftrightarrow 10x^2-16x+4=x^2-29x \\ \Leftrightarrow 9x^2+13x+4=0 \\\text{We zoeken de oplossingen van } \color{blue}{9x^2+13x+4=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (13)^2-4.9.4 & &\\ & = 169-144 & & \\ & = 25 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-13-\sqrt25}{2.9} & & = \frac{-13+\sqrt25}{2.9} \\ & = \frac{-18}{18} & & = \frac{-8}{18} \\ & = -1 & & = \frac{-4}{9} \\ \\ V &= \Big\{ -1 ; \frac{-4}{9} \Big\} & &\end{align} \\ -----------------\)
  3. \(\frac{4}{3}x^2+\frac{13}{9}x+\frac{1}{3}=0\\ \Leftrightarrow \color{red}{9.} \left(\frac{4}{3}x^2+\frac{13}{9}x+\frac{1}{3}\right)=0 \color{red}{.9} \\ \Leftrightarrow 12x^2+13x+3=0 \\\text{We zoeken de oplossingen van } \color{blue}{12x^2+13x+3=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (13)^2-4.12.3 & &\\ & = 169-144 & & \\ & = 25 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-13-\sqrt25}{2.12} & & = \frac{-13+\sqrt25}{2.12} \\ & = \frac{-18}{24} & & = \frac{-8}{24} \\ & = \frac{-3}{4} & & = \frac{-1}{3} \\ \\ V &= \Big\{ \frac{-3}{4} ; \frac{-1}{3} \Big\} & &\end{align} \\ -----------------\)
  4. \(x(9x+23)=3(x-12) \\ \Leftrightarrow 9x^2+23x=3x-36 \\ \Leftrightarrow 9x^2+20x+36=0 \\\text{We zoeken de oplossingen van } \color{blue}{9x^2+20x+36=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (20)^2-4.9.36 & &\\ & = 400-1296 & & \\ & = -896 & & \\ & < 0 \\V &= \varnothing \end{align} \\ -----------------\)
  5. \(-\frac{23}{4}x=-\frac{1}{4}x^2-33 \\ \Leftrightarrow \frac{1}{4}x^2-\frac{23}{4}x+33=0 \\ \Leftrightarrow \color{red}{4.} \left(\frac{1}{4}x^2-\frac{23}{4}x+33\right)=0 \color{red}{.4} \\ \Leftrightarrow x^2-23x+132=0 \\\text{We zoeken de oplossingen van } \color{blue}{x^2-23x+132=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (-23)^2-4.1.132 & &\\ & = 529-528 & & \\ & = 1 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-(-23)-\sqrt1}{2.1} & & = \frac{-(-23)+\sqrt1}{2.1} \\ & = \frac{22}{2} & & = \frac{24}{2} \\ & = 11 & & = 12 \\ \\ V &= \Big\{ 11 ; 12 \Big\} & &\end{align} \\ -----------------\)
  6. \(18x^2+\frac{25}{4}x+\frac{1}{2}=0\\ \Leftrightarrow \color{red}{4.} \left(18x^2+\frac{25}{4}x+\frac{1}{2}\right)=0 \color{red}{.4} \\ \Leftrightarrow 72x^2+25x+2=0 \\\text{We zoeken de oplossingen van } \color{blue}{72x^2+25x+2=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (25)^2-4.72.2 & &\\ & = 625-576 & & \\ & = 49 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-25-\sqrt49}{2.72} & & = \frac{-25+\sqrt49}{2.72} \\ & = \frac{-32}{144} & & = \frac{-18}{144} \\ & = \frac{-2}{9} & & = \frac{-1}{8} \\ \\ V &= \Big\{ \frac{-2}{9} ; \frac{-1}{8} \Big\} & &\end{align} \\ -----------------\)
  7. \((-5x+5)(2x+1)-x(-14x+38)=-95\\ \Leftrightarrow -10x^2-5x+10x+5 +14x^2-38x+95=0 \\ \Leftrightarrow 4x^2-38x+100=0 \\\text{We zoeken de oplossingen van } \color{blue}{4x^2-38x+100=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (-38)^2-4.4.100 & &\\ & = 1444-1600 & & \\ & = -156 & & \\ & < 0 \\V &= \varnothing \end{align} \\ -----------------\)
  8. \(17x^2-(14x-4)=x(x-6) \\ \Leftrightarrow 17x^2-14x+4=x^2-6x \\ \Leftrightarrow 16x^2-8x+4=0 \\\text{We zoeken de oplossingen van } \color{blue}{16x^2-8x+4=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (-8)^2-4.16.4 & &\\ & = 64-256 & & \\ & = -192 & & \\ & < 0 \\V &= \varnothing \end{align} \\ -----------------\)
  9. \((-5x+4)(-4x+5)-x(2x-10)=22\\ \Leftrightarrow 20x^2-25x-16x+20 -2x^2+10x-22=0 \\ \Leftrightarrow 18x^2+5x-2=0 \\\text{We zoeken de oplossingen van } \color{blue}{18x^2+5x-2=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (5)^2-4.18.(-2) & &\\ & = 25+144 & & \\ & = 169 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-5-\sqrt169}{2.18} & & = \frac{-5+\sqrt169}{2.18} \\ & = \frac{-18}{36} & & = \frac{8}{36} \\ & = \frac{-1}{2} & & = \frac{2}{9} \\ \\ V &= \Big\{ \frac{-1}{2} ; \frac{2}{9} \Big\} & &\end{align} \\ -----------------\)
  10. \(3x^2-(20x-36)=2x(x-12) \\ \Leftrightarrow 3x^2-20x+36=2x^2-24x \\ \Leftrightarrow x^2+4x+36=0 \\\text{We zoeken de oplossingen van } \color{blue}{x^2+4x+36=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (4)^2-4.1.36 & &\\ & = 16-144 & & \\ & = -128 & & \\ & < 0 \\V &= \varnothing \end{align} \\ -----------------\)
  11. \(-(6-41x)=-18x^2-(14-16x) \\ \Leftrightarrow -6+41x=-18x^2-14+16x \\ \Leftrightarrow 18x^2+25x+8=0 \\\text{We zoeken de oplossingen van } \color{blue}{18x^2+25x+8=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (25)^2-4.18.8 & &\\ & = 625-576 & & \\ & = 49 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-25-\sqrt49}{2.18} & & = \frac{-25+\sqrt49}{2.18} \\ & = \frac{-32}{36} & & = \frac{-18}{36} \\ & = \frac{-8}{9} & & = \frac{-1}{2} \\ \\ V &= \Big\{ \frac{-8}{9} ; \frac{-1}{2} \Big\} & &\end{align} \\ -----------------\)
  12. \(-(11-8x)=-12x^2-(8-3x) \\ \Leftrightarrow -11+8x=-12x^2-8+3x \\ \Leftrightarrow 12x^2+5x-3=0 \\\text{We zoeken de oplossingen van } \color{blue}{12x^2+5x-3=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (5)^2-4.12.(-3) & &\\ & = 25+144 & & \\ & = 169 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-5-\sqrt169}{2.12} & & = \frac{-5+\sqrt169}{2.12} \\ & = \frac{-18}{24} & & = \frac{8}{24} \\ & = \frac{-3}{4} & & = \frac{1}{3} \\ \\ V &= \Big\{ \frac{-3}{4} ; \frac{1}{3} \Big\} & &\end{align} \\ -----------------\)
Oefeningengenerator vanhoeckes.be/wiskunde 2024-12-03 18:28:26