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Gebruik de discriminant om volgende vierkantsvergelijkingen op te lossen

  1. \(2x^2-(10x-8)=x(x-4)\)
  2. \(\frac{1}{3}x^2+\frac{13}{3}x+12=0\)
  3. \((2x-4)(4x+3)-x(7x+8)=-52\)
  4. \((x-1)(4x+5)-x(-44x-7)=-2\)
  5. \(-\frac{1}{3}x=-\frac{1}{3}x^2+44\)
  6. \(x(x+22)=16(x+1)\)
  7. \(\frac{5}{8}x=-\frac{1}{4}x^2+\frac{9}{4}\)
  8. \((-4x-5)(x-4)-x(-5x+28)=5\)
  9. \((-x-2)(x-4)-x(-2x+6)=3\)
  10. \((-4x+1)(3x+4)-x(-13x-14)=-60\)
  11. \(-(14-3x)=-x^2-(38-17x)\)
  12. \(x^2+\frac{3}{4}x-\frac{1}{4}=0\)

Gebruik de discriminant om volgende vierkantsvergelijkingen op te lossen

Verbetersleutel

  1. \(2x^2-(10x-8)=x(x-4) \\ \Leftrightarrow 2x^2-10x+8=x^2-4x \\ \Leftrightarrow x^2-6x+8=0 \\\text{We zoeken de oplossingen van } \color{blue}{x^2-6x+8=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (-6)^2-4.1.8 & &\\ & = 36-32 & & \\ & = 4 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-(-6)-\sqrt4}{2.1} & & = \frac{-(-6)+\sqrt4}{2.1} \\ & = \frac{4}{2} & & = \frac{8}{2} \\ & = 2 & & = 4 \\ \\ V &= \Big\{ 2 ; 4 \Big\} & &\end{align} \\ -----------------\)
  2. \(\frac{1}{3}x^2+\frac{13}{3}x+12=0\\ \Leftrightarrow \color{red}{3.} \left(\frac{1}{3}x^2+\frac{13}{3}x+12\right)=0 \color{red}{.3} \\ \Leftrightarrow x^2+13x+36=0 \\\text{We zoeken de oplossingen van } \color{blue}{x^2+13x+36=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (13)^2-4.1.36 & &\\ & = 169-144 & & \\ & = 25 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-13-\sqrt25}{2.1} & & = \frac{-13+\sqrt25}{2.1} \\ & = \frac{-18}{2} & & = \frac{-8}{2} \\ & = -9 & & = -4 \\ \\ V &= \Big\{ -9 ; -4 \Big\} & &\end{align} \\ -----------------\)
  3. \((2x-4)(4x+3)-x(7x+8)=-52\\ \Leftrightarrow 8x^2+6x-16x-12 -7x^2-8x+52=0 \\ \Leftrightarrow x^2-14x+40=0 \\\text{We zoeken de oplossingen van } \color{blue}{x^2-14x+40=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (-14)^2-4.1.40 & &\\ & = 196-160 & & \\ & = 36 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-(-14)-\sqrt36}{2.1} & & = \frac{-(-14)+\sqrt36}{2.1} \\ & = \frac{8}{2} & & = \frac{20}{2} \\ & = 4 & & = 10 \\ \\ V &= \Big\{ 4 ; 10 \Big\} & &\end{align} \\ -----------------\)
  4. \((x-1)(4x+5)-x(-44x-7)=-2\\ \Leftrightarrow 4x^2+5x-4x-5 +44x^2+7x+2=0 \\ \Leftrightarrow 48x^2+7x-3=0 \\\text{We zoeken de oplossingen van } \color{blue}{48x^2+7x-3=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (7)^2-4.48.(-3) & &\\ & = 49+576 & & \\ & = 625 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-7-\sqrt625}{2.48} & & = \frac{-7+\sqrt625}{2.48} \\ & = \frac{-32}{96} & & = \frac{18}{96} \\ & = \frac{-1}{3} & & = \frac{3}{16} \\ \\ V &= \Big\{ \frac{-1}{3} ; \frac{3}{16} \Big\} & &\end{align} \\ -----------------\)
  5. \(-\frac{1}{3}x=-\frac{1}{3}x^2+44 \\ \Leftrightarrow \frac{1}{3}x^2-\frac{1}{3}x-44=0 \\ \Leftrightarrow \color{red}{3.} \left(\frac{1}{3}x^2-\frac{1}{3}x-44\right)=0 \color{red}{.3} \\ \Leftrightarrow x^2-x-132=0 \\\text{We zoeken de oplossingen van } \color{blue}{x^2-x-132=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (-1)^2-4.1.(-132) & &\\ & = 1+528 & & \\ & = 529 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-(-1)-\sqrt529}{2.1} & & = \frac{-(-1)+\sqrt529}{2.1} \\ & = \frac{-22}{2} & & = \frac{24}{2} \\ & = -11 & & = 12 \\ \\ V &= \Big\{ -11 ; 12 \Big\} & &\end{align} \\ -----------------\)
  6. \(x(x+22)=16(x+1) \\ \Leftrightarrow x^2+22x=16x+16 \\ \Leftrightarrow x^2+6x-16=0 \\\text{We zoeken de oplossingen van } \color{blue}{x^2+6x-16=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (6)^2-4.1.(-16) & &\\ & = 36+64 & & \\ & = 100 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-6-\sqrt100}{2.1} & & = \frac{-6+\sqrt100}{2.1} \\ & = \frac{-16}{2} & & = \frac{4}{2} \\ & = -8 & & = 2 \\ \\ V &= \Big\{ -8 ; 2 \Big\} & &\end{align} \\ -----------------\)
  7. \(\frac{5}{8}x=-\frac{1}{4}x^2+\frac{9}{4} \\ \Leftrightarrow \frac{1}{4}x^2+\frac{5}{8}x-\frac{9}{4}=0 \\ \Leftrightarrow \color{red}{8.} \left(\frac{1}{4}x^2+\frac{5}{8}x-\frac{9}{4}\right)=0 \color{red}{.8} \\ \Leftrightarrow 2x^2+5x-18=0 \\\text{We zoeken de oplossingen van } \color{blue}{2x^2+5x-18=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (5)^2-4.2.(-18) & &\\ & = 25+144 & & \\ & = 169 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-5-\sqrt169}{2.2} & & = \frac{-5+\sqrt169}{2.2} \\ & = \frac{-18}{4} & & = \frac{8}{4} \\ & = \frac{-9}{2} & & = 2 \\ \\ V &= \Big\{ \frac{-9}{2} ; 2 \Big\} & &\end{align} \\ -----------------\)
  8. \((-4x-5)(x-4)-x(-5x+28)=5\\ \Leftrightarrow -4x^2+16x-5x+20 +5x^2-28x-5=0 \\ \Leftrightarrow x^2+8x+15=0 \\\text{We zoeken de oplossingen van } \color{blue}{x^2+8x+15=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (8)^2-4.1.15 & &\\ & = 64-60 & & \\ & = 4 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-8-\sqrt4}{2.1} & & = \frac{-8+\sqrt4}{2.1} \\ & = \frac{-10}{2} & & = \frac{-6}{2} \\ & = -5 & & = -3 \\ \\ V &= \Big\{ -5 ; -3 \Big\} & &\end{align} \\ -----------------\)
  9. \((-x-2)(x-4)-x(-2x+6)=3\\ \Leftrightarrow -x^2+4x-2x+8 +2x^2-6x-3=0 \\ \Leftrightarrow x^2+6x+5=0 \\\text{We zoeken de oplossingen van } \color{blue}{x^2+6x+5=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (6)^2-4.1.5 & &\\ & = 36-20 & & \\ & = 16 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-6-\sqrt16}{2.1} & & = \frac{-6+\sqrt16}{2.1} \\ & = \frac{-10}{2} & & = \frac{-2}{2} \\ & = -5 & & = -1 \\ \\ V &= \Big\{ -5 ; -1 \Big\} & &\end{align} \\ -----------------\)
  10. \((-4x+1)(3x+4)-x(-13x-14)=-60\\ \Leftrightarrow -12x^2-16x+3x+4 +13x^2+14x+60=0 \\ \Leftrightarrow x^2+2x+64=0 \\\text{We zoeken de oplossingen van } \color{blue}{x^2+2x+64=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (2)^2-4.1.64 & &\\ & = 4-256 & & \\ & = -252 & & \\ & < 0 \\V &= \varnothing \end{align} \\ -----------------\)
  11. \(-(14-3x)=-x^2-(38-17x) \\ \Leftrightarrow -14+3x=-x^2-38+17x \\ \Leftrightarrow x^2-14x+24=0 \\\text{We zoeken de oplossingen van } \color{blue}{x^2-14x+24=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (-14)^2-4.1.24 & &\\ & = 196-96 & & \\ & = 100 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-(-14)-\sqrt100}{2.1} & & = \frac{-(-14)+\sqrt100}{2.1} \\ & = \frac{4}{2} & & = \frac{24}{2} \\ & = 2 & & = 12 \\ \\ V &= \Big\{ 2 ; 12 \Big\} & &\end{align} \\ -----------------\)
  12. \(x^2+\frac{3}{4}x-\frac{1}{4}=0\\ \Leftrightarrow \color{red}{4.} \left(x^2+\frac{3}{4}x-\frac{1}{4}\right)=0 \color{red}{.4} \\ \Leftrightarrow 4x^2+3x-1=0 \\\text{We zoeken de oplossingen van } \color{blue}{4x^2+3x-1=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (3)^2-4.4.(-1) & &\\ & = 9+16 & & \\ & = 25 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-3-\sqrt25}{2.4} & & = \frac{-3+\sqrt25}{2.4} \\ & = \frac{-8}{8} & & = \frac{2}{8} \\ & = -1 & & = \frac{1}{4} \\ \\ V &= \Big\{ -1 ; \frac{1}{4} \Big\} & &\end{align} \\ -----------------\)
Oefeningengenerator wiskundeoefeningen.be 2026-03-31 13:55:23
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