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Gebruik de discriminant om volgende vierkantsvergelijkingen op te lossen

1. \(3x^2-\frac{22}{3}x+\frac{25}{3}=0\)
2. \(\frac{36}{5}x^2+\frac{5}{2}x+\frac{1}{5}=0\)
3. \(-\frac{5}{4}x=-\frac{1}{2}x^2-\frac{25}{32}\)
4. \(2x^2-(18x-50)=x(x-33)\)
5. \(\frac{8}{3}x^2+x-\frac{1}{6}=0\)
6. \((-4x+4)(-2x-3)-x(7x-20)=-112\)
7. \(7x^2-(14x-4)=6x(x-3)\)
8. \(\frac{5}{6}x=-\frac{1}{5}x^2-\frac{4}{5}\)
9. \(3x^2-(16x-7)=2x(x-4)\)
10. \(\frac{1}{18}x^2+\frac{1}{2}x-2=0\)
11. \(\frac{1}{5}x^2-\frac{6}{5}x+\frac{49}{5}=0\)
12. \(x(24x+28)=3(x-2)\)

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Gebruik de discriminant om volgende vierkantsvergelijkingen op te lossen

Verbetersleutel

1. \(3x^2-\frac{22}{3}x+\frac{25}{3}=0\\ \Leftrightarrow \color{red}{3.} \left(3x^2-\frac{22}{3}x+\frac{25}{3}\right)=0 \color{red}{.3} \\ \Leftrightarrow 9x^2-22x+25=0 \\\text{We zoeken de oplossingen van } \color{blue}{9x^2-22x+25=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (-22)^2-4.9.25 & &\\ & = 484-900 & & \\ & = -416 & & \\ & < 0 \\V &= \varnothing \end{align} \\ -----------------\)
2. \(\frac{36}{5}x^2+\frac{5}{2}x+\frac{1}{5}=0\\ \Leftrightarrow \color{red}{10.} \left(\frac{36}{5}x^2+\frac{5}{2}x+\frac{1}{5}\right)=0 \color{red}{.10} \\ \Leftrightarrow 72x^2+25x+2=0 \\\text{We zoeken de oplossingen van } \color{blue}{72x^2+25x+2=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (25)^2-4.72.2 & &\\ & = 625-576 & & \\ & = 49 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-25-\sqrt49}{2.72} & & = \frac{-25+\sqrt49}{2.72} \\ & = \frac{-32}{144} & & = \frac{-18}{144} \\ & = \frac{-2}{9} & & = \frac{-1}{8} \\ \\ V &= \Big\{ \frac{-2}{9} ; \frac{-1}{8} \Big\} & &\end{align} \\ -----------------\)
3. \(-\frac{5}{4}x=-\frac{1}{2}x^2-\frac{25}{32} \\ \Leftrightarrow \frac{1}{2}x^2-\frac{5}{4}x+\frac{25}{32}=0 \\ \Leftrightarrow \color{red}{32.} \left(\frac{1}{2}x^2-\frac{5}{4}x+\frac{25}{32}\right)=0 \color{red}{.32} \\ \Leftrightarrow 16x^2-40x+25=0 \\\text{We zoeken de oplossingen van } \color{blue}{16x^2-40x+25=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (-40)^2-4.16.25 & &\\ & = 1600-1600 & & \\ & = 0 & & \\ x & = \frac{-b\pm \sqrt{D}}{2.a} & & \\ & = \frac{-(-40)}{2.16} & & \\ & = \frac{5}{4} & & \\V &= \Big\{ \frac{5}{4} \Big\} & &\end{align} \\ -----------------\)
4. \(2x^2-(18x-50)=x(x-33) \\ \Leftrightarrow 2x^2-18x+50=x^2-33x \\ \Leftrightarrow x^2+15x+50=0 \\\text{We zoeken de oplossingen van } \color{blue}{x^2+15x+50=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (15)^2-4.1.50 & &\\ & = 225-200 & & \\ & = 25 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-15-\sqrt25}{2.1} & & = \frac{-15+\sqrt25}{2.1} \\ & = \frac{-20}{2} & & = \frac{-10}{2} \\ & = -10 & & = -5 \\ \\ V &= \Big\{ -10 ; -5 \Big\} & &\end{align} \\ -----------------\)
5. \(\frac{8}{3}x^2+x-\frac{1}{6}=0\\ \Leftrightarrow \color{red}{6.} \left(\frac{8}{3}x^2+x-\frac{1}{6}\right)=0 \color{red}{.6} \\ \Leftrightarrow 16x^2+6x-1=0 \\\text{We zoeken de oplossingen van } \color{blue}{16x^2+6x-1=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (6)^2-4.16.(-1) & &\\ & = 36+64 & & \\ & = 100 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-6-\sqrt100}{2.16} & & = \frac{-6+\sqrt100}{2.16} \\ & = \frac{-16}{32} & & = \frac{4}{32} \\ & = \frac{-1}{2} & & = \frac{1}{8} \\ \\ V &= \Big\{ \frac{-1}{2} ; \frac{1}{8} \Big\} & &\end{align} \\ -----------------\)
6. \((-4x+4)(-2x-3)-x(7x-20)=-112\\ \Leftrightarrow 8x^2+12x-8x-12 -7x^2+20x+112=0 \\ \Leftrightarrow x^2+20x+100=0 \\\text{We zoeken de oplossingen van } \color{blue}{x^2+20x+100=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (20)^2-4.1.100 & &\\ & = 400-400 & & \\ & = 0 & & \\ x & = \frac{-b\pm \sqrt{D}}{2.a} & & \\ & = \frac{-20}{2.1} & & \\ & = -10 & & \\V &= \Big\{ -10 \Big\} & &\end{align} \\ -----------------\)
7. \(7x^2-(14x-4)=6x(x-3) \\ \Leftrightarrow 7x^2-14x+4=6x^2-18x \\ \Leftrightarrow x^2+4x+4=0 \\\text{We zoeken de oplossingen van } \color{blue}{x^2+4x+4=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (4)^2-4.1.4 & &\\ & = 16-16 & & \\ & = 0 & & \\ x & = \frac{-b\pm \sqrt{D}}{2.a} & & \\ & = \frac{-4}{2.1} & & \\ & = -2 & & \\V &= \Big\{ -2 \Big\} & &\end{align} \\ -----------------\)
8. \(\frac{5}{6}x=-\frac{1}{5}x^2-\frac{4}{5} \\ \Leftrightarrow \frac{1}{5}x^2+\frac{5}{6}x+\frac{4}{5}=0 \\ \Leftrightarrow \color{red}{30.} \left(\frac{1}{5}x^2+\frac{5}{6}x+\frac{4}{5}\right)=0 \color{red}{.30} \\ \Leftrightarrow 6x^2+25x+24=0 \\\text{We zoeken de oplossingen van } \color{blue}{6x^2+25x+24=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (25)^2-4.6.24 & &\\ & = 625-576 & & \\ & = 49 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-25-\sqrt49}{2.6} & & = \frac{-25+\sqrt49}{2.6} \\ & = \frac{-32}{12} & & = \frac{-18}{12} \\ & = \frac{-8}{3} & & = \frac{-3}{2} \\ \\ V &= \Big\{ \frac{-8}{3} ; \frac{-3}{2} \Big\} & &\end{align} \\ -----------------\)
9. \(3x^2-(16x-7)=2x(x-4) \\ \Leftrightarrow 3x^2-16x+7=2x^2-8x \\ \Leftrightarrow x^2-8x+7=0 \\\text{We zoeken de oplossingen van } \color{blue}{x^2-8x+7=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (-8)^2-4.1.7 & &\\ & = 64-28 & & \\ & = 36 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-(-8)-\sqrt36}{2.1} & & = \frac{-(-8)+\sqrt36}{2.1} \\ & = \frac{2}{2} & & = \frac{14}{2} \\ & = 1 & & = 7 \\ \\ V &= \Big\{ 1 ; 7 \Big\} & &\end{align} \\ -----------------\)
10. \(\frac{1}{18}x^2+\frac{1}{2}x-2=0\\ \Leftrightarrow \color{red}{18.} \left(\frac{1}{18}x^2+\frac{1}{2}x-2\right)=0 \color{red}{.18} \\ \Leftrightarrow x^2+9x-36=0 \\\text{We zoeken de oplossingen van } \color{blue}{x^2+9x-36=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (9)^2-4.1.(-36) & &\\ & = 81+144 & & \\ & = 225 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-9-\sqrt225}{2.1} & & = \frac{-9+\sqrt225}{2.1} \\ & = \frac{-24}{2} & & = \frac{6}{2} \\ & = -12 & & = 3 \\ \\ V &= \Big\{ -12 ; 3 \Big\} & &\end{align} \\ -----------------\)
11. \(\frac{1}{5}x^2-\frac{6}{5}x+\frac{49}{5}=0\\ \Leftrightarrow \color{red}{5.} \left(\frac{1}{5}x^2-\frac{6}{5}x+\frac{49}{5}\right)=0 \color{red}{.5} \\ \Leftrightarrow x^2-6x+49=0 \\\text{We zoeken de oplossingen van } \color{blue}{x^2-6x+49=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (-6)^2-4.1.49 & &\\ & = 36-196 & & \\ & = -160 & & \\ & < 0 \\V &= \varnothing \end{align} \\ -----------------\)
12. \(x(24x+28)=3(x-2) \\ \Leftrightarrow 24x^2+28x=3x-6 \\ \Leftrightarrow 24x^2+25x+6=0 \\\text{We zoeken de oplossingen van } \color{blue}{24x^2+25x+6=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (25)^2-4.24.6 & &\\ & = 625-576 & & \\ & = 49 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-25-\sqrt49}{2.24} & & = \frac{-25+\sqrt49}{2.24} \\ & = \frac{-32}{48} & & = \frac{-18}{48} \\ & = \frac{-2}{3} & & = \frac{-3}{8} \\ \\ V &= \Big\{ \frac{-2}{3} ; \frac{-3}{8} \Big\} & &\end{align} \\ -----------------\)