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Gebruik de discriminant om volgende vierkantsvergelijkingen op te lossen

1. \(-(4-53x)=-9x^2-(68-5x)\)
2. \(-(9-16x)=-2x^2-(-63-9x)\)
3. \(\frac{13}{3}x=-\frac{4}{3}x^2-3\)
4. \(\frac{1}{10}x^2+\frac{1}{5}x-\frac{12}{5}=0\)
5. \((4x+2)(x+5)-x(2x+25)=28\)
6. \(28x^2-(5x+6)=4x(x-3)\)
7. \(2x^2-(9x+55)=x(x-3)\)
8. \(\frac{17}{4}x=-\frac{1}{4}x^2-\frac{35}{2}\)
9. \(-(11-20x)=-16x^2-(15-4x)\)
10. \(\frac{1}{8}x^2+\frac{5}{2}x+12=0\)
11. \(2x^2-(19x-32)=x(x-7)\)
12. \(-\frac{3}{4}x=-\frac{1}{4}x^2+1\)

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Gebruik de discriminant om volgende vierkantsvergelijkingen op te lossen

Verbetersleutel

1. \(-(4-53x)=-9x^2-(68-5x) \\ \Leftrightarrow -4+53x=-9x^2-68+5x \\ \Leftrightarrow 9x^2+48x+64=0 \\\text{We zoeken de oplossingen van } \color{blue}{9x^2+48x+64=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (48)^2-4.9.64 & &\\ & = 2304-2304 & & \\ & = 0 & & \\ x & = \frac{-b\pm \sqrt{D}}{2.a} & & \\ & = \frac{-48}{2.9} & & \\ & = -\frac{8}{3} & & \\V &= \Big\{ -\frac{8}{3} \Big\} & &\end{align} \\ -----------------\)
2. \(-(9-16x)=-2x^2-(-63-9x) \\ \Leftrightarrow -9+16x=-2x^2+63+9x \\ \Leftrightarrow 2x^2+7x-72=0 \\\text{We zoeken de oplossingen van } \color{blue}{2x^2+7x-72=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (7)^2-4.2.(-72) & &\\ & = 49+576 & & \\ & = 625 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-7-\sqrt625}{2.2} & & = \frac{-7+\sqrt625}{2.2} \\ & = \frac{-32}{4} & & = \frac{18}{4} \\ & = -8 & & = \frac{9}{2} \\ \\ V &= \Big\{ -8 ; \frac{9}{2} \Big\} & &\end{align} \\ -----------------\)
3. \(\frac{13}{3}x=-\frac{4}{3}x^2-3 \\ \Leftrightarrow \frac{4}{3}x^2+\frac{13}{3}x+3=0 \\ \Leftrightarrow \color{red}{3.} \left(\frac{4}{3}x^2+\frac{13}{3}x+3\right)=0 \color{red}{.3} \\ \Leftrightarrow 4x^2+13x+9=0 \\\text{We zoeken de oplossingen van } \color{blue}{4x^2+13x+9=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (13)^2-4.4.9 & &\\ & = 169-144 & & \\ & = 25 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-13-\sqrt25}{2.4} & & = \frac{-13+\sqrt25}{2.4} \\ & = \frac{-18}{8} & & = \frac{-8}{8} \\ & = \frac{-9}{4} & & = -1 \\ \\ V &= \Big\{ \frac{-9}{4} ; -1 \Big\} & &\end{align} \\ -----------------\)
4. \(\frac{1}{10}x^2+\frac{1}{5}x-\frac{12}{5}=0\\ \Leftrightarrow \color{red}{10.} \left(\frac{1}{10}x^2+\frac{1}{5}x-\frac{12}{5}\right)=0 \color{red}{.10} \\ \Leftrightarrow x^2+2x-24=0 \\\text{We zoeken de oplossingen van } \color{blue}{x^2+2x-24=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (2)^2-4.1.(-24) & &\\ & = 4+96 & & \\ & = 100 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-2-\sqrt100}{2.1} & & = \frac{-2+\sqrt100}{2.1} \\ & = \frac{-12}{2} & & = \frac{8}{2} \\ & = -6 & & = 4 \\ \\ V &= \Big\{ -6 ; 4 \Big\} & &\end{align} \\ -----------------\)
5. \((4x+2)(x+5)-x(2x+25)=28\\ \Leftrightarrow 4x^2+20x+2x+10 -2x^2-25x-28=0 \\ \Leftrightarrow 2x^2+5x-18=0 \\\text{We zoeken de oplossingen van } \color{blue}{2x^2+5x-18=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (5)^2-4.2.(-18) & &\\ & = 25+144 & & \\ & = 169 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-5-\sqrt169}{2.2} & & = \frac{-5+\sqrt169}{2.2} \\ & = \frac{-18}{4} & & = \frac{8}{4} \\ & = \frac{-9}{2} & & = 2 \\ \\ V &= \Big\{ \frac{-9}{2} ; 2 \Big\} & &\end{align} \\ -----------------\)
6. \(28x^2-(5x+6)=4x(x-3) \\ \Leftrightarrow 28x^2-5x-6=4x^2-12x \\ \Leftrightarrow 24x^2+7x-6=0 \\\text{We zoeken de oplossingen van } \color{blue}{24x^2+7x-6=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (7)^2-4.24.(-6) & &\\ & = 49+576 & & \\ & = 625 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-7-\sqrt625}{2.24} & & = \frac{-7+\sqrt625}{2.24} \\ & = \frac{-32}{48} & & = \frac{18}{48} \\ & = \frac{-2}{3} & & = \frac{3}{8} \\ \\ V &= \Big\{ \frac{-2}{3} ; \frac{3}{8} \Big\} & &\end{align} \\ -----------------\)
7. \(2x^2-(9x+55)=x(x-3) \\ \Leftrightarrow 2x^2-9x-55=x^2-3x \\ \Leftrightarrow x^2-6x-55=0 \\\text{We zoeken de oplossingen van } \color{blue}{x^2-6x-55=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (-6)^2-4.1.(-55) & &\\ & = 36+220 & & \\ & = 256 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-(-6)-\sqrt256}{2.1} & & = \frac{-(-6)+\sqrt256}{2.1} \\ & = \frac{-10}{2} & & = \frac{22}{2} \\ & = -5 & & = 11 \\ \\ V &= \Big\{ -5 ; 11 \Big\} & &\end{align} \\ -----------------\)
8. \(\frac{17}{4}x=-\frac{1}{4}x^2-\frac{35}{2} \\ \Leftrightarrow \frac{1}{4}x^2+\frac{17}{4}x+\frac{35}{2}=0 \\ \Leftrightarrow \color{red}{4.} \left(\frac{1}{4}x^2+\frac{17}{4}x+\frac{35}{2}\right)=0 \color{red}{.4} \\ \Leftrightarrow x^2+17x+70=0 \\\text{We zoeken de oplossingen van } \color{blue}{x^2+17x+70=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (17)^2-4.1.70 & &\\ & = 289-280 & & \\ & = 9 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-17-\sqrt9}{2.1} & & = \frac{-17+\sqrt9}{2.1} \\ & = \frac{-20}{2} & & = \frac{-14}{2} \\ & = -10 & & = -7 \\ \\ V &= \Big\{ -10 ; -7 \Big\} & &\end{align} \\ -----------------\)
9. \(-(11-20x)=-16x^2-(15-4x) \\ \Leftrightarrow -11+20x=-16x^2-15+4x \\ \Leftrightarrow 16x^2+16x+4=0 \\\text{We zoeken de oplossingen van } \color{blue}{16x^2+16x+4=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (16)^2-4.16.4 & &\\ & = 256-256 & & \\ & = 0 & & \\ x & = \frac{-b\pm \sqrt{D}}{2.a} & & \\ & = \frac{-16}{2.16} & & \\ & = -\frac{1}{2} & & \\V &= \Big\{ -\frac{1}{2} \Big\} & &\end{align} \\ -----------------\)
10. \(\frac{1}{8}x^2+\frac{5}{2}x+12=0\\ \Leftrightarrow \color{red}{8.} \left(\frac{1}{8}x^2+\frac{5}{2}x+12\right)=0 \color{red}{.8} \\ \Leftrightarrow x^2+20x+96=0 \\\text{We zoeken de oplossingen van } \color{blue}{x^2+20x+96=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (20)^2-4.1.96 & &\\ & = 400-384 & & \\ & = 16 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-20-\sqrt16}{2.1} & & = \frac{-20+\sqrt16}{2.1} \\ & = \frac{-24}{2} & & = \frac{-16}{2} \\ & = -12 & & = -8 \\ \\ V &= \Big\{ -12 ; -8 \Big\} & &\end{align} \\ -----------------\)
11. \(2x^2-(19x-32)=x(x-7) \\ \Leftrightarrow 2x^2-19x+32=x^2-7x \\ \Leftrightarrow x^2-12x+32=0 \\\text{We zoeken de oplossingen van } \color{blue}{x^2-12x+32=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (-12)^2-4.1.32 & &\\ & = 144-128 & & \\ & = 16 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-(-12)-\sqrt16}{2.1} & & = \frac{-(-12)+\sqrt16}{2.1} \\ & = \frac{8}{2} & & = \frac{16}{2} \\ & = 4 & & = 8 \\ \\ V &= \Big\{ 4 ; 8 \Big\} & &\end{align} \\ -----------------\)
12. \(-\frac{3}{4}x=-\frac{1}{4}x^2+1 \\ \Leftrightarrow \frac{1}{4}x^2-\frac{3}{4}x-1=0 \\ \Leftrightarrow \color{red}{4.} \left(\frac{1}{4}x^2-\frac{3}{4}x-1\right)=0 \color{red}{.4} \\ \Leftrightarrow x^2-3x-4=0 \\\text{We zoeken de oplossingen van } \color{blue}{x^2-3x-4=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (-3)^2-4.1.(-4) & &\\ & = 9+16 & & \\ & = 25 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-(-3)-\sqrt25}{2.1} & & = \frac{-(-3)+\sqrt25}{2.1} \\ & = \frac{-2}{2} & & = \frac{8}{2} \\ & = -1 & & = 4 \\ \\ V &= \Big\{ -1 ; 4 \Big\} & &\end{align} \\ -----------------\)