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Gebruik de discriminant om volgende vierkantsvergelijkingen op te lossen

  1. \(5x^2-(7x-49)=x(x-35)\)
  2. \(x(6x+7)=-6(x+1)\)
  3. \(x(x+34)=40(x+1)\)
  4. \(-\frac{1}{4}x=-\frac{1}{2}x^2-\frac{1}{32}\)
  5. \(x(4x-77)=-49(x+1)\)
  6. \(4x^2-(8x+12)=x(x-13)\)
  7. \(20x^2-(9x-48)=17x(x-2)\)
  8. \(\frac{17}{16}x=-\frac{1}{4}x^2-\frac{1}{4}\)
  9. \(\frac{1}{4}x^2+x+\frac{49}{4}=0\)
  10. \(-\frac{1}{5}x=-\frac{1}{15}x^2+\frac{36}{5}\)
  11. \(\frac{1}{2}x=-\frac{1}{4}x^2+\frac{63}{4}\)
  12. \(\frac{5}{4}x=-\frac{1}{16}x^2-6\)

Gebruik de discriminant om volgende vierkantsvergelijkingen op te lossen

Verbetersleutel

  1. \(5x^2-(7x-49)=x(x-35) \\ \Leftrightarrow 5x^2-7x+49=x^2-35x \\ \Leftrightarrow 4x^2+28x+49=0 \\\text{We zoeken de oplossingen van } \color{blue}{4x^2+28x+49=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (28)^2-4.4.49 & &\\ & = 784-784 & & \\ & = 0 & & \\ x & = \frac{-b\pm \sqrt{D}}{2.a} & & \\ & = \frac{-28}{2.4} & & \\ & = -\frac{7}{2} & & \\V &= \Big\{ -\frac{7}{2} \Big\} & &\end{align} \\ -----------------\)
  2. \(x(6x+7)=-6(x+1) \\ \Leftrightarrow 6x^2+7x=-6x-6 \\ \Leftrightarrow 6x^2+13x+6=0 \\\text{We zoeken de oplossingen van } \color{blue}{6x^2+13x+6=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (13)^2-4.6.6 & &\\ & = 169-144 & & \\ & = 25 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-13-\sqrt25}{2.6} & & = \frac{-13+\sqrt25}{2.6} \\ & = \frac{-18}{12} & & = \frac{-8}{12} \\ & = \frac{-3}{2} & & = \frac{-2}{3} \\ \\ V &= \Big\{ \frac{-3}{2} ; \frac{-2}{3} \Big\} & &\end{align} \\ -----------------\)
  3. \(x(x+34)=40(x+1) \\ \Leftrightarrow x^2+34x=40x+40 \\ \Leftrightarrow x^2-6x-40=0 \\\text{We zoeken de oplossingen van } \color{blue}{x^2-6x-40=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (-6)^2-4.1.(-40) & &\\ & = 36+160 & & \\ & = 196 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-(-6)-\sqrt196}{2.1} & & = \frac{-(-6)+\sqrt196}{2.1} \\ & = \frac{-8}{2} & & = \frac{20}{2} \\ & = -4 & & = 10 \\ \\ V &= \Big\{ -4 ; 10 \Big\} & &\end{align} \\ -----------------\)
  4. \(-\frac{1}{4}x=-\frac{1}{2}x^2-\frac{1}{32} \\ \Leftrightarrow \frac{1}{2}x^2-\frac{1}{4}x+\frac{1}{32}=0 \\ \Leftrightarrow \color{red}{32.} \left(\frac{1}{2}x^2-\frac{1}{4}x+\frac{1}{32}\right)=0 \color{red}{.32} \\ \Leftrightarrow 16x^2-8x+1=0 \\\text{We zoeken de oplossingen van } \color{blue}{16x^2-8x+1=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (-8)^2-4.16.1 & &\\ & = 64-64 & & \\ & = 0 & & \\ x & = \frac{-b\pm \sqrt{D}}{2.a} & & \\ & = \frac{-(-8)}{2.16} & & \\ & = \frac{1}{4} & & \\V &= \Big\{ \frac{1}{4} \Big\} & &\end{align} \\ -----------------\)
  5. \(x(4x-77)=-49(x+1) \\ \Leftrightarrow 4x^2-77x=-49x-49 \\ \Leftrightarrow 4x^2-28x+49=0 \\\text{We zoeken de oplossingen van } \color{blue}{4x^2-28x+49=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (-28)^2-4.4.49 & &\\ & = 784-784 & & \\ & = 0 & & \\ x & = \frac{-b\pm \sqrt{D}}{2.a} & & \\ & = \frac{-(-28)}{2.4} & & \\ & = \frac{7}{2} & & \\V &= \Big\{ \frac{7}{2} \Big\} & &\end{align} \\ -----------------\)
  6. \(4x^2-(8x+12)=x(x-13) \\ \Leftrightarrow 4x^2-8x-12=x^2-13x \\ \Leftrightarrow 3x^2+5x-12=0 \\\text{We zoeken de oplossingen van } \color{blue}{3x^2+5x-12=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (5)^2-4.3.(-12) & &\\ & = 25+144 & & \\ & = 169 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-5-\sqrt169}{2.3} & & = \frac{-5+\sqrt169}{2.3} \\ & = \frac{-18}{6} & & = \frac{8}{6} \\ & = -3 & & = \frac{4}{3} \\ \\ V &= \Big\{ -3 ; \frac{4}{3} \Big\} & &\end{align} \\ -----------------\)
  7. \(20x^2-(9x-48)=17x(x-2) \\ \Leftrightarrow 20x^2-9x+48=17x^2-34x \\ \Leftrightarrow 3x^2+25x+48=0 \\\text{We zoeken de oplossingen van } \color{blue}{3x^2+25x+48=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (25)^2-4.3.48 & &\\ & = 625-576 & & \\ & = 49 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-25-\sqrt49}{2.3} & & = \frac{-25+\sqrt49}{2.3} \\ & = \frac{-32}{6} & & = \frac{-18}{6} \\ & = \frac{-16}{3} & & = -3 \\ \\ V &= \Big\{ \frac{-16}{3} ; -3 \Big\} & &\end{align} \\ -----------------\)
  8. \(\frac{17}{16}x=-\frac{1}{4}x^2-\frac{1}{4} \\ \Leftrightarrow \frac{1}{4}x^2+\frac{17}{16}x+\frac{1}{4}=0 \\ \Leftrightarrow \color{red}{16.} \left(\frac{1}{4}x^2+\frac{17}{16}x+\frac{1}{4}\right)=0 \color{red}{.16} \\ \Leftrightarrow 4x^2+17x+4=0 \\\text{We zoeken de oplossingen van } \color{blue}{4x^2+17x+4=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (17)^2-4.4.4 & &\\ & = 289-64 & & \\ & = 225 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-17-\sqrt225}{2.4} & & = \frac{-17+\sqrt225}{2.4} \\ & = \frac{-32}{8} & & = \frac{-2}{8} \\ & = -4 & & = \frac{-1}{4} \\ \\ V &= \Big\{ -4 ; \frac{-1}{4} \Big\} & &\end{align} \\ -----------------\)
  9. \(\frac{1}{4}x^2+x+\frac{49}{4}=0\\ \Leftrightarrow \color{red}{4.} \left(\frac{1}{4}x^2+x+\frac{49}{4}\right)=0 \color{red}{.4} \\ \Leftrightarrow x^2+4x+49=0 \\\text{We zoeken de oplossingen van } \color{blue}{x^2+4x+49=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (4)^2-4.1.49 & &\\ & = 16-196 & & \\ & = -180 & & \\ & < 0 \\V &= \varnothing \end{align} \\ -----------------\)
  10. \(-\frac{1}{5}x=-\frac{1}{15}x^2+\frac{36}{5} \\ \Leftrightarrow \frac{1}{15}x^2-\frac{1}{5}x-\frac{36}{5}=0 \\ \Leftrightarrow \color{red}{15.} \left(\frac{1}{15}x^2-\frac{1}{5}x-\frac{36}{5}\right)=0 \color{red}{.15} \\ \Leftrightarrow x^2-3x-108=0 \\\text{We zoeken de oplossingen van } \color{blue}{x^2-3x-108=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (-3)^2-4.1.(-108) & &\\ & = 9+432 & & \\ & = 441 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-(-3)-\sqrt441}{2.1} & & = \frac{-(-3)+\sqrt441}{2.1} \\ & = \frac{-18}{2} & & = \frac{24}{2} \\ & = -9 & & = 12 \\ \\ V &= \Big\{ -9 ; 12 \Big\} & &\end{align} \\ -----------------\)
  11. \(\frac{1}{2}x=-\frac{1}{4}x^2+\frac{63}{4} \\ \Leftrightarrow \frac{1}{4}x^2+\frac{1}{2}x-\frac{63}{4}=0 \\ \Leftrightarrow \color{red}{4.} \left(\frac{1}{4}x^2+\frac{1}{2}x-\frac{63}{4}\right)=0 \color{red}{.4} \\ \Leftrightarrow x^2+2x-63=0 \\\text{We zoeken de oplossingen van } \color{blue}{x^2+2x-63=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (2)^2-4.1.(-63) & &\\ & = 4+252 & & \\ & = 256 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-2-\sqrt256}{2.1} & & = \frac{-2+\sqrt256}{2.1} \\ & = \frac{-18}{2} & & = \frac{14}{2} \\ & = -9 & & = 7 \\ \\ V &= \Big\{ -9 ; 7 \Big\} & &\end{align} \\ -----------------\)
  12. \(\frac{5}{4}x=-\frac{1}{16}x^2-6 \\ \Leftrightarrow \frac{1}{16}x^2+\frac{5}{4}x+6=0 \\ \Leftrightarrow \color{red}{16.} \left(\frac{1}{16}x^2+\frac{5}{4}x+6\right)=0 \color{red}{.16} \\ \Leftrightarrow x^2+20x+96=0 \\\text{We zoeken de oplossingen van } \color{blue}{x^2+20x+96=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (20)^2-4.1.96 & &\\ & = 400-384 & & \\ & = 16 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-20-\sqrt16}{2.1} & & = \frac{-20+\sqrt16}{2.1} \\ & = \frac{-24}{2} & & = \frac{-16}{2} \\ & = -12 & & = -8 \\ \\ V &= \Big\{ -12 ; -8 \Big\} & &\end{align} \\ -----------------\)
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