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Gebruik de discriminant om volgende vierkantsvergelijkingen op te lossen

  1. \(\frac{1}{2}x=-\frac{1}{16}x^2-1\)
  2. \(\frac{1}{4}x^2+\frac{9}{4}x-\frac{5}{2}=0\)
  3. \(-(12+87x)=-16x^2-(156-9x)\)
  4. \((-4x+3)(x-2)-x(-8x-5)=30\)
  5. \(-(15+2x)=-x^2-(64-2x)\)
  6. \(x(x-5)=2(x-6)\)
  7. \((2x+4)(4x-4)-x(-4x-37)=-19\)
  8. \((-x+3)(5x-2)-x(-14x+12)=-31\)
  9. \((4x-1)(-x-4)-x(-6x-27)=12\)
  10. \(\frac{1}{8}x^2+\frac{5}{4}x+\frac{9}{2}=0\)
  11. \(\frac{1}{5}x^2+\frac{7}{30}x-\frac{4}{5}=0\)
  12. \(10x^2-(18x+4)=x(x-23)\)

Gebruik de discriminant om volgende vierkantsvergelijkingen op te lossen

Verbetersleutel

  1. \(\frac{1}{2}x=-\frac{1}{16}x^2-1 \\ \Leftrightarrow \frac{1}{16}x^2+\frac{1}{2}x+1=0 \\ \Leftrightarrow \color{red}{16.} \left(\frac{1}{16}x^2+\frac{1}{2}x+1\right)=0 \color{red}{.16} \\ \Leftrightarrow 9x^2+72x+144=0 \\\text{We zoeken de oplossingen van } \color{blue}{9x^2+72x+144=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (72)^2-4.9.144 & &\\ & = 5184-5184 & & \\ & = 0 & & \\ x & = \frac{-b\pm \sqrt{D}}{2.a} & & \\ & = \frac{-72}{2.9} & & \\ & = -4 & & \\V &= \Big\{ -4 \Big\} & &\end{align} \\ -----------------\)
  2. \(\frac{1}{4}x^2+\frac{9}{4}x-\frac{5}{2}=0\\ \Leftrightarrow \color{red}{4.} \left(\frac{1}{4}x^2+\frac{9}{4}x-\frac{5}{2}\right)=0 \color{red}{.4} \\ \Leftrightarrow x^2+9x-10=0 \\\text{We zoeken de oplossingen van } \color{blue}{x^2+9x-10=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (9)^2-4.1.(-10) & &\\ & = 81+40 & & \\ & = 121 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-9-\sqrt121}{2.1} & & = \frac{-9+\sqrt121}{2.1} \\ & = \frac{-20}{2} & & = \frac{2}{2} \\ & = -10 & & = 1 \\ \\ V &= \Big\{ -10 ; 1 \Big\} & &\end{align} \\ -----------------\)
  3. \(-(12+87x)=-16x^2-(156-9x) \\ \Leftrightarrow -12-87x=-16x^2-156+9x \\ \Leftrightarrow 16x^2-96x+144=0 \\\text{We zoeken de oplossingen van } \color{blue}{16x^2-96x+144=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (-96)^2-4.16.144 & &\\ & = 9216-9216 & & \\ & = 0 & & \\ x & = \frac{-b\pm \sqrt{D}}{2.a} & & \\ & = \frac{-(-96)}{2.16} & & \\ & = 3 & & \\V &= \Big\{ 3 \Big\} & &\end{align} \\ -----------------\)
  4. \((-4x+3)(x-2)-x(-8x-5)=30\\ \Leftrightarrow -4x^2+8x+3x-6 +8x^2+5x-30=0 \\ \Leftrightarrow 4x^2+7x-36=0 \\\text{We zoeken de oplossingen van } \color{blue}{4x^2+7x-36=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (7)^2-4.4.(-36) & &\\ & = 49+576 & & \\ & = 625 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-7-\sqrt625}{2.4} & & = \frac{-7+\sqrt625}{2.4} \\ & = \frac{-32}{8} & & = \frac{18}{8} \\ & = -4 & & = \frac{9}{4} \\ \\ V &= \Big\{ -4 ; \frac{9}{4} \Big\} & &\end{align} \\ -----------------\)
  5. \(-(15+2x)=-x^2-(64-2x) \\ \Leftrightarrow -15-2x=-x^2-64+2x \\ \Leftrightarrow x^2-4x+49=0 \\\text{We zoeken de oplossingen van } \color{blue}{x^2-4x+49=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (-4)^2-4.1.49 & &\\ & = 16-196 & & \\ & = -180 & & \\ & < 0 \\V &= \varnothing \end{align} \\ -----------------\)
  6. \(x(x-5)=2(x-6) \\ \Leftrightarrow x^2-5x=2x-12 \\ \Leftrightarrow x^2-7x+12=0 \\\text{We zoeken de oplossingen van } \color{blue}{x^2-7x+12=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (-7)^2-4.1.12 & &\\ & = 49-48 & & \\ & = 1 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-(-7)-\sqrt1}{2.1} & & = \frac{-(-7)+\sqrt1}{2.1} \\ & = \frac{6}{2} & & = \frac{8}{2} \\ & = 3 & & = 4 \\ \\ V &= \Big\{ 3 ; 4 \Big\} & &\end{align} \\ -----------------\)
  7. \((2x+4)(4x-4)-x(-4x-37)=-19\\ \Leftrightarrow 8x^2-8x+16x-16 +4x^2+37x+19=0 \\ \Leftrightarrow 12x^2+13x+3=0 \\\text{We zoeken de oplossingen van } \color{blue}{12x^2+13x+3=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (13)^2-4.12.3 & &\\ & = 169-144 & & \\ & = 25 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-13-\sqrt25}{2.12} & & = \frac{-13+\sqrt25}{2.12} \\ & = \frac{-18}{24} & & = \frac{-8}{24} \\ & = \frac{-3}{4} & & = \frac{-1}{3} \\ \\ V &= \Big\{ \frac{-3}{4} ; \frac{-1}{3} \Big\} & &\end{align} \\ -----------------\)
  8. \((-x+3)(5x-2)-x(-14x+12)=-31\\ \Leftrightarrow -5x^2+2x+15x-6 +14x^2-12x+31=0 \\ \Leftrightarrow 9x^2-16x+25=0 \\\text{We zoeken de oplossingen van } \color{blue}{9x^2-16x+25=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (-16)^2-4.9.25 & &\\ & = 256-900 & & \\ & = -644 & & \\ & < 0 \\V &= \varnothing \end{align} \\ -----------------\)
  9. \((4x-1)(-x-4)-x(-6x-27)=12\\ \Leftrightarrow -4x^2-16x+x+4 +6x^2+27x-12=0 \\ \Leftrightarrow 2x^2+15x-8=0 \\\text{We zoeken de oplossingen van } \color{blue}{2x^2+15x-8=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (15)^2-4.2.(-8) & &\\ & = 225+64 & & \\ & = 289 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-15-\sqrt289}{2.2} & & = \frac{-15+\sqrt289}{2.2} \\ & = \frac{-32}{4} & & = \frac{2}{4} \\ & = -8 & & = \frac{1}{2} \\ \\ V &= \Big\{ -8 ; \frac{1}{2} \Big\} & &\end{align} \\ -----------------\)
  10. \(\frac{1}{8}x^2+\frac{5}{4}x+\frac{9}{2}=0\\ \Leftrightarrow \color{red}{8.} \left(\frac{1}{8}x^2+\frac{5}{4}x+\frac{9}{2}\right)=0 \color{red}{.8} \\ \Leftrightarrow x^2+10x+36=0 \\\text{We zoeken de oplossingen van } \color{blue}{x^2+10x+36=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (10)^2-4.1.36 & &\\ & = 100-144 & & \\ & = -44 & & \\ & < 0 \\V &= \varnothing \end{align} \\ -----------------\)
  11. \(\frac{1}{5}x^2+\frac{7}{30}x-\frac{4}{5}=0\\ \Leftrightarrow \color{red}{30.} \left(\frac{1}{5}x^2+\frac{7}{30}x-\frac{4}{5}\right)=0 \color{red}{.30} \\ \Leftrightarrow 6x^2+7x-24=0 \\\text{We zoeken de oplossingen van } \color{blue}{6x^2+7x-24=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (7)^2-4.6.(-24) & &\\ & = 49+576 & & \\ & = 625 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-7-\sqrt625}{2.6} & & = \frac{-7+\sqrt625}{2.6} \\ & = \frac{-32}{12} & & = \frac{18}{12} \\ & = \frac{-8}{3} & & = \frac{3}{2} \\ \\ V &= \Big\{ \frac{-8}{3} ; \frac{3}{2} \Big\} & &\end{align} \\ -----------------\)
  12. \(10x^2-(18x+4)=x(x-23) \\ \Leftrightarrow 10x^2-18x-4=x^2-23x \\ \Leftrightarrow 9x^2+5x-4=0 \\\text{We zoeken de oplossingen van } \color{blue}{9x^2+5x-4=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (5)^2-4.9.(-4) & &\\ & = 25+144 & & \\ & = 169 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-5-\sqrt169}{2.9} & & = \frac{-5+\sqrt169}{2.9} \\ & = \frac{-18}{18} & & = \frac{8}{18} \\ & = -1 & & = \frac{4}{9} \\ \\ V &= \Big\{ -1 ; \frac{4}{9} \Big\} & &\end{align} \\ -----------------\)
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