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Gebruik de discriminant om volgende vierkantsvergelijkingen op te lossen

1. \(x(x-8)=6(x-4)\)
2. \(28x^2-(8x-25)=19x(x-2)\)
3. \(-(11-10x)=-9x^2-(15-16x)\)
4. \(-(15-11x)=-x^2-(29-20x)\)
5. \(-(9-42x)=-3x^2-(57-17x)\)
6. \(\frac{1}{3}x^2+x+\frac{3}{4}=0\)
7. \(x(9x+0)=-36(x+1)\)
8. \(\frac{9}{2}x^2-2x+\frac{1}{2}=0\)
9. \(x(x+33)=11(x-11)\)
10. \(\frac{1}{6}x^2-x+\frac{50}{3}=0\)
11. \((-4x+4)(-5x-1)-x(19x-21)=-112\)
12. \(\frac{9}{5}x^2+\frac{5}{4}x+\frac{1}{5}=0\)

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Gebruik de discriminant om volgende vierkantsvergelijkingen op te lossen

Verbetersleutel

1. \(x(x-8)=6(x-4) \\ \Leftrightarrow x^2-8x=6x-24 \\ \Leftrightarrow x^2-14x+24=0 \\\text{We zoeken de oplossingen van } \color{blue}{x^2-14x+24=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (-14)^2-4.1.24 & &\\ & = 196-96 & & \\ & = 100 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-(-14)-\sqrt100}{2.1} & & = \frac{-(-14)+\sqrt100}{2.1} \\ & = \frac{4}{2} & & = \frac{24}{2} \\ & = 2 & & = 12 \\ \\ V &= \Big\{ 2 ; 12 \Big\} & &\end{align} \\ -----------------\)
2. \(28x^2-(8x-25)=19x(x-2) \\ \Leftrightarrow 28x^2-8x+25=19x^2-38x \\ \Leftrightarrow 9x^2+30x+25=0 \\\text{We zoeken de oplossingen van } \color{blue}{9x^2+30x+25=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (30)^2-4.9.25 & &\\ & = 900-900 & & \\ & = 0 & & \\ x & = \frac{-b\pm \sqrt{D}}{2.a} & & \\ & = \frac{-30}{2.9} & & \\ & = -\frac{5}{3} & & \\V &= \Big\{ -\frac{5}{3} \Big\} & &\end{align} \\ -----------------\)
3. \(-(11-10x)=-9x^2-(15-16x) \\ \Leftrightarrow -11+10x=-9x^2-15+16x \\ \Leftrightarrow 9x^2-6x+4=0 \\\text{We zoeken de oplossingen van } \color{blue}{9x^2-6x+4=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (-6)^2-4.9.4 & &\\ & = 36-144 & & \\ & = -108 & & \\ & < 0 \\V &= \varnothing \end{align} \\ -----------------\)
4. \(-(15-11x)=-x^2-(29-20x) \\ \Leftrightarrow -15+11x=-x^2-29+20x \\ \Leftrightarrow x^2-9x+14=0 \\\text{We zoeken de oplossingen van } \color{blue}{x^2-9x+14=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (-9)^2-4.1.14 & &\\ & = 81-56 & & \\ & = 25 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-(-9)-\sqrt25}{2.1} & & = \frac{-(-9)+\sqrt25}{2.1} \\ & = \frac{4}{2} & & = \frac{14}{2} \\ & = 2 & & = 7 \\ \\ V &= \Big\{ 2 ; 7 \Big\} & &\end{align} \\ -----------------\)
5. \(-(9-42x)=-3x^2-(57-17x) \\ \Leftrightarrow -9+42x=-3x^2-57+17x \\ \Leftrightarrow 3x^2+25x+48=0 \\\text{We zoeken de oplossingen van } \color{blue}{3x^2+25x+48=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (25)^2-4.3.48 & &\\ & = 625-576 & & \\ & = 49 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-25-\sqrt49}{2.3} & & = \frac{-25+\sqrt49}{2.3} \\ & = \frac{-32}{6} & & = \frac{-18}{6} \\ & = \frac{-16}{3} & & = -3 \\ \\ V &= \Big\{ \frac{-16}{3} ; -3 \Big\} & &\end{align} \\ -----------------\)
6. \(\frac{1}{3}x^2+x+\frac{3}{4}=0\\ \Leftrightarrow \color{red}{12.} \left(\frac{1}{3}x^2+x+\frac{3}{4}\right)=0 \color{red}{.12} \\ \Leftrightarrow 4x^2+12x+9=0 \\\text{We zoeken de oplossingen van } \color{blue}{4x^2+12x+9=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (12)^2-4.4.9 & &\\ & = 144-144 & & \\ & = 0 & & \\ x & = \frac{-b\pm \sqrt{D}}{2.a} & & \\ & = \frac{-12}{2.4} & & \\ & = -\frac{3}{2} & & \\V &= \Big\{ -\frac{3}{2} \Big\} & &\end{align} \\ -----------------\)
7. \(x(9x+0)=-36(x+1) \\ \Leftrightarrow 9x^2+0x=-36x-36 \\ \Leftrightarrow 9x^2+36x+36=0 \\\text{We zoeken de oplossingen van } \color{blue}{9x^2+36x+36=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (36)^2-4.9.36 & &\\ & = 1296-1296 & & \\ & = 0 & & \\ x & = \frac{-b\pm \sqrt{D}}{2.a} & & \\ & = \frac{-36}{2.9} & & \\ & = -2 & & \\V &= \Big\{ -2 \Big\} & &\end{align} \\ -----------------\)
8. \(\frac{9}{2}x^2-2x+\frac{1}{2}=0\\ \Leftrightarrow \color{red}{2.} \left(\frac{9}{2}x^2-2x+\frac{1}{2}\right)=0 \color{red}{.2} \\ \Leftrightarrow 9x^2-4x+1=0 \\\text{We zoeken de oplossingen van } \color{blue}{9x^2-4x+1=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (-4)^2-4.9.1 & &\\ & = 16-36 & & \\ & = -20 & & \\ & < 0 \\V &= \varnothing \end{align} \\ -----------------\)
9. \(x(x+33)=11(x-11) \\ \Leftrightarrow x^2+33x=11x-121 \\ \Leftrightarrow x^2+22x+121=0 \\\text{We zoeken de oplossingen van } \color{blue}{x^2+22x+121=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (22)^2-4.1.121 & &\\ & = 484-484 & & \\ & = 0 & & \\ x & = \frac{-b\pm \sqrt{D}}{2.a} & & \\ & = \frac{-22}{2.1} & & \\ & = -11 & & \\V &= \Big\{ -11 \Big\} & &\end{align} \\ -----------------\)
10. \(\frac{1}{6}x^2-x+\frac{50}{3}=0\\ \Leftrightarrow \color{red}{6.} \left(\frac{1}{6}x^2-x+\frac{50}{3}\right)=0 \color{red}{.6} \\ \Leftrightarrow x^2-6x+100=0 \\\text{We zoeken de oplossingen van } \color{blue}{x^2-6x+100=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (-6)^2-4.1.100 & &\\ & = 36-400 & & \\ & = -364 & & \\ & < 0 \\V &= \varnothing \end{align} \\ -----------------\)
11. \((-4x+4)(-5x-1)-x(19x-21)=-112\\ \Leftrightarrow 20x^2+4x-20x-4 -19x^2+21x+112=0 \\ \Leftrightarrow x^2+21x+108=0 \\\text{We zoeken de oplossingen van } \color{blue}{x^2+21x+108=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (21)^2-4.1.108 & &\\ & = 441-432 & & \\ & = 9 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-21-\sqrt9}{2.1} & & = \frac{-21+\sqrt9}{2.1} \\ & = \frac{-24}{2} & & = \frac{-18}{2} \\ & = -12 & & = -9 \\ \\ V &= \Big\{ -12 ; -9 \Big\} & &\end{align} \\ -----------------\)
12. \(\frac{9}{5}x^2+\frac{5}{4}x+\frac{1}{5}=0\\ \Leftrightarrow \color{red}{20.} \left(\frac{9}{5}x^2+\frac{5}{4}x+\frac{1}{5}\right)=0 \color{red}{.20} \\ \Leftrightarrow 36x^2+25x+4=0 \\\text{We zoeken de oplossingen van } \color{blue}{36x^2+25x+4=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (25)^2-4.36.4 & &\\ & = 625-576 & & \\ & = 49 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-25-\sqrt49}{2.36} & & = \frac{-25+\sqrt49}{2.36} \\ & = \frac{-32}{72} & & = \frac{-18}{72} \\ & = \frac{-4}{9} & & = \frac{-1}{4} \\ \\ V &= \Big\{ \frac{-4}{9} ; \frac{-1}{4} \Big\} & &\end{align} \\ -----------------\)