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Gebruik de discriminant om volgende vierkantsvergelijkingen op te lossen

1. \(x(x+23)=6(x-11)\)
2. \(17x^2-(15x-1)=x(x-13)\)
3. \(x(x+34)=24(x+1)\)
4. \(-\frac{1}{3}x=-\frac{1}{12}x^2-\frac{1}{3}\)
5. \(\frac{1}{2}x^2+\frac{1}{2}x+\frac{9}{32}=0\)
6. \(\frac{5}{6}x=-\frac{1}{3}x^2-\frac{1}{3}\)
7. \(x(x+67)=60(x+1)\)
8. \(-(5-27x)=-18x^2-(13-2x)\)
9. \((-x-5)(x+1)-x(-17x+66)=-86\)
10. \(-(8-19x)=-x^2-(56-3x)\)
11. \(x=-\frac{3}{20}x^2-\frac{5}{3}\)
12. \(\frac{1}{5}x^2-\frac{9}{5}x-\frac{22}{5}=0\)

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Gebruik de discriminant om volgende vierkantsvergelijkingen op te lossen

Verbetersleutel

1. \(x(x+23)=6(x-11) \\ \Leftrightarrow x^2+23x=6x-66 \\ \Leftrightarrow x^2+17x+66=0 \\\text{We zoeken de oplossingen van } \color{blue}{x^2+17x+66=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (17)^2-4.1.66 & &\\ & = 289-264 & & \\ & = 25 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-17-\sqrt25}{2.1} & & = \frac{-17+\sqrt25}{2.1} \\ & = \frac{-22}{2} & & = \frac{-12}{2} \\ & = -11 & & = -6 \\ \\ V &= \Big\{ -11 ; -6 \Big\} & &\end{align} \\ -----------------\)
2. \(17x^2-(15x-1)=x(x-13) \\ \Leftrightarrow 17x^2-15x+1=x^2-13x \\ \Leftrightarrow 16x^2-2x+1=0 \\\text{We zoeken de oplossingen van } \color{blue}{16x^2-2x+1=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (-2)^2-4.16.1 & &\\ & = 4-64 & & \\ & = -60 & & \\ & < 0 \\V &= \varnothing \end{align} \\ -----------------\)
3. \(x(x+34)=24(x+1) \\ \Leftrightarrow x^2+34x=24x+24 \\ \Leftrightarrow x^2+10x-24=0 \\\text{We zoeken de oplossingen van } \color{blue}{x^2+10x-24=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (10)^2-4.1.(-24) & &\\ & = 100+96 & & \\ & = 196 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-10-\sqrt196}{2.1} & & = \frac{-10+\sqrt196}{2.1} \\ & = \frac{-24}{2} & & = \frac{4}{2} \\ & = -12 & & = 2 \\ \\ V &= \Big\{ -12 ; 2 \Big\} & &\end{align} \\ -----------------\)
4. \(-\frac{1}{3}x=-\frac{1}{12}x^2-\frac{1}{3} \\ \Leftrightarrow \frac{1}{12}x^2-\frac{1}{3}x+\frac{1}{3}=0 \\ \Leftrightarrow \color{red}{12.} \left(\frac{1}{12}x^2-\frac{1}{3}x+\frac{1}{3}\right)=0 \color{red}{.12} \\ \Leftrightarrow 16x^2-64x+64=0 \\\text{We zoeken de oplossingen van } \color{blue}{16x^2-64x+64=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (-64)^2-4.16.64 & &\\ & = 4096-4096 & & \\ & = 0 & & \\ x & = \frac{-b\pm \sqrt{D}}{2.a} & & \\ & = \frac{-(-64)}{2.16} & & \\ & = 2 & & \\V &= \Big\{ 2 \Big\} & &\end{align} \\ -----------------\)
5. \(\frac{1}{2}x^2+\frac{1}{2}x+\frac{9}{32}=0\\ \Leftrightarrow \color{red}{32.} \left(\frac{1}{2}x^2+\frac{1}{2}x+\frac{9}{32}\right)=0 \color{red}{.32} \\ \Leftrightarrow 16x^2+16x+9=0 \\\text{We zoeken de oplossingen van } \color{blue}{16x^2+16x+9=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (16)^2-4.16.9 & &\\ & = 256-576 & & \\ & = -320 & & \\ & < 0 \\V &= \varnothing \end{align} \\ -----------------\)
6. \(\frac{5}{6}x=-\frac{1}{3}x^2-\frac{1}{3} \\ \Leftrightarrow \frac{1}{3}x^2+\frac{5}{6}x+\frac{1}{3}=0 \\ \Leftrightarrow \color{red}{6.} \left(\frac{1}{3}x^2+\frac{5}{6}x+\frac{1}{3}\right)=0 \color{red}{.6} \\ \Leftrightarrow 2x^2+5x+2=0 \\\text{We zoeken de oplossingen van } \color{blue}{2x^2+5x+2=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (5)^2-4.2.2 & &\\ & = 25-16 & & \\ & = 9 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-5-\sqrt9}{2.2} & & = \frac{-5+\sqrt9}{2.2} \\ & = \frac{-8}{4} & & = \frac{-2}{4} \\ & = -2 & & = \frac{-1}{2} \\ \\ V &= \Big\{ -2 ; \frac{-1}{2} \Big\} & &\end{align} \\ -----------------\)
7. \(x(x+67)=60(x+1) \\ \Leftrightarrow x^2+67x=60x+60 \\ \Leftrightarrow x^2+7x-60=0 \\\text{We zoeken de oplossingen van } \color{blue}{x^2+7x-60=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (7)^2-4.1.(-60) & &\\ & = 49+240 & & \\ & = 289 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-7-\sqrt289}{2.1} & & = \frac{-7+\sqrt289}{2.1} \\ & = \frac{-24}{2} & & = \frac{10}{2} \\ & = -12 & & = 5 \\ \\ V &= \Big\{ -12 ; 5 \Big\} & &\end{align} \\ -----------------\)
8. \(-(5-27x)=-18x^2-(13-2x) \\ \Leftrightarrow -5+27x=-18x^2-13+2x \\ \Leftrightarrow 18x^2+25x+8=0 \\\text{We zoeken de oplossingen van } \color{blue}{18x^2+25x+8=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (25)^2-4.18.8 & &\\ & = 625-576 & & \\ & = 49 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-25-\sqrt49}{2.18} & & = \frac{-25+\sqrt49}{2.18} \\ & = \frac{-32}{36} & & = \frac{-18}{36} \\ & = \frac{-8}{9} & & = \frac{-1}{2} \\ \\ V &= \Big\{ \frac{-8}{9} ; \frac{-1}{2} \Big\} & &\end{align} \\ -----------------\)
9. \((-x-5)(x+1)-x(-17x+66)=-86\\ \Leftrightarrow -x^2-x-5x-5 +17x^2-66x+86=0 \\ \Leftrightarrow 16x^2-72x+81=0 \\\text{We zoeken de oplossingen van } \color{blue}{16x^2-72x+81=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (-72)^2-4.16.81 & &\\ & = 5184-5184 & & \\ & = 0 & & \\ x & = \frac{-b\pm \sqrt{D}}{2.a} & & \\ & = \frac{-(-72)}{2.16} & & \\ & = \frac{9}{4} & & \\V &= \Big\{ \frac{9}{4} \Big\} & &\end{align} \\ -----------------\)
10. \(-(8-19x)=-x^2-(56-3x) \\ \Leftrightarrow -8+19x=-x^2-56+3x \\ \Leftrightarrow x^2+16x+48=0 \\\text{We zoeken de oplossingen van } \color{blue}{x^2+16x+48=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (16)^2-4.1.48 & &\\ & = 256-192 & & \\ & = 64 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-16-\sqrt64}{2.1} & & = \frac{-16+\sqrt64}{2.1} \\ & = \frac{-24}{2} & & = \frac{-8}{2} \\ & = -12 & & = -4 \\ \\ V &= \Big\{ -12 ; -4 \Big\} & &\end{align} \\ -----------------\)
11. \(x=-\frac{3}{20}x^2-\frac{5}{3} \\ \Leftrightarrow \frac{3}{20}x^2+x+\frac{5}{3}=0 \\ \Leftrightarrow \color{red}{60.} \left(\frac{3}{20}x^2+x+\frac{5}{3}\right)=0 \color{red}{.60} \\ \Leftrightarrow 9x^2+60x+100=0 \\\text{We zoeken de oplossingen van } \color{blue}{9x^2+60x+100=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (60)^2-4.9.100 & &\\ & = 3600-3600 & & \\ & = 0 & & \\ x & = \frac{-b\pm \sqrt{D}}{2.a} & & \\ & = \frac{-60}{2.9} & & \\ & = -\frac{10}{3} & & \\V &= \Big\{ -\frac{10}{3} \Big\} & &\end{align} \\ -----------------\)
12. \(\frac{1}{5}x^2-\frac{9}{5}x-\frac{22}{5}=0\\ \Leftrightarrow \color{red}{5.} \left(\frac{1}{5}x^2-\frac{9}{5}x-\frac{22}{5}\right)=0 \color{red}{.5} \\ \Leftrightarrow x^2-9x-22=0 \\\text{We zoeken de oplossingen van } \color{blue}{x^2-9x-22=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (-9)^2-4.1.(-22) & &\\ & = 81+88 & & \\ & = 169 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-(-9)-\sqrt169}{2.1} & & = \frac{-(-9)+\sqrt169}{2.1} \\ & = \frac{-4}{2} & & = \frac{22}{2} \\ & = -2 & & = 11 \\ \\ V &= \Big\{ -2 ; 11 \Big\} & &\end{align} \\ -----------------\)