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Gebruik de discriminant om volgende vierkantsvergelijkingen op te lossen

1. \((-3x+2)(x-1)-x(-19x+89)=-123\)
2. \(\frac{1}{2}x=-\frac{1}{8}x^2+\frac{3}{2}\)
3. \(5x^2-(12x-16)=x(x+4)\)
4. \(\frac{1}{2}x^2+\frac{13}{2}x+18=0\)
5. \(\frac{1}{2}x^2-x+\frac{121}{32}=0\)
6. \((-3x+1)(4x+3)-x(-13x-12)=-2\)
7. \(2x^2-(5x+120)=x(x-3)\)
8. \(\frac{1}{5}x^2-\frac{2}{5}x+5=0\)
9. \(10x^2-(9x-4)=x(x+3)\)
10. \(x(4x+19)=4(x+1)\)
11. \(-(7+7x)=-16x^2-(16-17x)\)
12. \(\frac{1}{24}x^2-\frac{1}{2}x+\frac{3}{2}=0\)

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Gebruik de discriminant om volgende vierkantsvergelijkingen op te lossen

Verbetersleutel

1. \((-3x+2)(x-1)-x(-19x+89)=-123\\ \Leftrightarrow -3x^2+3x+2x-2 +19x^2-89x+123=0 \\ \Leftrightarrow 16x^2-88x+121=0 \\\text{We zoeken de oplossingen van } \color{blue}{16x^2-88x+121=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (-88)^2-4.16.121 & &\\ & = 7744-7744 & & \\ & = 0 & & \\ x & = \frac{-b\pm \sqrt{D}}{2.a} & & \\ & = \frac{-(-88)}{2.16} & & \\ & = \frac{11}{4} & & \\V &= \Big\{ \frac{11}{4} \Big\} & &\end{align} \\ -----------------\)
2. \(\frac{1}{2}x=-\frac{1}{8}x^2+\frac{3}{2} \\ \Leftrightarrow \frac{1}{8}x^2+\frac{1}{2}x-\frac{3}{2}=0 \\ \Leftrightarrow \color{red}{8.} \left(\frac{1}{8}x^2+\frac{1}{2}x-\frac{3}{2}\right)=0 \color{red}{.8} \\ \Leftrightarrow x^2+4x-12=0 \\\text{We zoeken de oplossingen van } \color{blue}{x^2+4x-12=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (4)^2-4.1.(-12) & &\\ & = 16+48 & & \\ & = 64 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-4-\sqrt64}{2.1} & & = \frac{-4+\sqrt64}{2.1} \\ & = \frac{-12}{2} & & = \frac{4}{2} \\ & = -6 & & = 2 \\ \\ V &= \Big\{ -6 ; 2 \Big\} & &\end{align} \\ -----------------\)
3. \(5x^2-(12x-16)=x(x+4) \\ \Leftrightarrow 5x^2-12x+16=x^2+4x \\ \Leftrightarrow 4x^2-16x+16=0 \\\text{We zoeken de oplossingen van } \color{blue}{4x^2-16x+16=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (-16)^2-4.4.16 & &\\ & = 256-256 & & \\ & = 0 & & \\ x & = \frac{-b\pm \sqrt{D}}{2.a} & & \\ & = \frac{-(-16)}{2.4} & & \\ & = 2 & & \\V &= \Big\{ 2 \Big\} & &\end{align} \\ -----------------\)
4. \(\frac{1}{2}x^2+\frac{13}{2}x+18=0\\ \Leftrightarrow \color{red}{2.} \left(\frac{1}{2}x^2+\frac{13}{2}x+18\right)=0 \color{red}{.2} \\ \Leftrightarrow x^2+13x+36=0 \\\text{We zoeken de oplossingen van } \color{blue}{x^2+13x+36=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (13)^2-4.1.36 & &\\ & = 169-144 & & \\ & = 25 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-13-\sqrt25}{2.1} & & = \frac{-13+\sqrt25}{2.1} \\ & = \frac{-18}{2} & & = \frac{-8}{2} \\ & = -9 & & = -4 \\ \\ V &= \Big\{ -9 ; -4 \Big\} & &\end{align} \\ -----------------\)
5. \(\frac{1}{2}x^2-x+\frac{121}{32}=0\\ \Leftrightarrow \color{red}{32.} \left(\frac{1}{2}x^2-x+\frac{121}{32}\right)=0 \color{red}{.32} \\ \Leftrightarrow 16x^2-32x+121=0 \\\text{We zoeken de oplossingen van } \color{blue}{16x^2-32x+121=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (-32)^2-4.16.121 & &\\ & = 1024-7744 & & \\ & = -6720 & & \\ & < 0 \\V &= \varnothing \end{align} \\ -----------------\)
6. \((-3x+1)(4x+3)-x(-13x-12)=-2\\ \Leftrightarrow -12x^2-9x+4x+3 +13x^2+12x+2=0 \\ \Leftrightarrow x^2+6x+5=0 \\\text{We zoeken de oplossingen van } \color{blue}{x^2+6x+5=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (6)^2-4.1.5 & &\\ & = 36-20 & & \\ & = 16 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-6-\sqrt16}{2.1} & & = \frac{-6+\sqrt16}{2.1} \\ & = \frac{-10}{2} & & = \frac{-2}{2} \\ & = -5 & & = -1 \\ \\ V &= \Big\{ -5 ; -1 \Big\} & &\end{align} \\ -----------------\)
7. \(2x^2-(5x+120)=x(x-3) \\ \Leftrightarrow 2x^2-5x-120=x^2-3x \\ \Leftrightarrow x^2-2x-120=0 \\\text{We zoeken de oplossingen van } \color{blue}{x^2-2x-120=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (-2)^2-4.1.(-120) & &\\ & = 4+480 & & \\ & = 484 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-(-2)-\sqrt484}{2.1} & & = \frac{-(-2)+\sqrt484}{2.1} \\ & = \frac{-20}{2} & & = \frac{24}{2} \\ & = -10 & & = 12 \\ \\ V &= \Big\{ -10 ; 12 \Big\} & &\end{align} \\ -----------------\)
8. \(\frac{1}{5}x^2-\frac{2}{5}x+5=0\\ \Leftrightarrow \color{red}{5.} \left(\frac{1}{5}x^2-\frac{2}{5}x+5\right)=0 \color{red}{.5} \\ \Leftrightarrow x^2-2x+25=0 \\\text{We zoeken de oplossingen van } \color{blue}{x^2-2x+25=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (-2)^2-4.1.25 & &\\ & = 4-100 & & \\ & = -96 & & \\ & < 0 \\V &= \varnothing \end{align} \\ -----------------\)
9. \(10x^2-(9x-4)=x(x+3) \\ \Leftrightarrow 10x^2-9x+4=x^2+3x \\ \Leftrightarrow 9x^2-12x+4=0 \\\text{We zoeken de oplossingen van } \color{blue}{9x^2-12x+4=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (-12)^2-4.9.4 & &\\ & = 144-144 & & \\ & = 0 & & \\ x & = \frac{-b\pm \sqrt{D}}{2.a} & & \\ & = \frac{-(-12)}{2.9} & & \\ & = \frac{2}{3} & & \\V &= \Big\{ \frac{2}{3} \Big\} & &\end{align} \\ -----------------\)
10. \(x(4x+19)=4(x+1) \\ \Leftrightarrow 4x^2+19x=4x+4 \\ \Leftrightarrow 4x^2+15x-4=0 \\\text{We zoeken de oplossingen van } \color{blue}{4x^2+15x-4=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (15)^2-4.4.(-4) & &\\ & = 225+64 & & \\ & = 289 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-15-\sqrt289}{2.4} & & = \frac{-15+\sqrt289}{2.4} \\ & = \frac{-32}{8} & & = \frac{2}{8} \\ & = -4 & & = \frac{1}{4} \\ \\ V &= \Big\{ -4 ; \frac{1}{4} \Big\} & &\end{align} \\ -----------------\)
11. \(-(7+7x)=-16x^2-(16-17x) \\ \Leftrightarrow -7-7x=-16x^2-16+17x \\ \Leftrightarrow 16x^2-24x+9=0 \\\text{We zoeken de oplossingen van } \color{blue}{16x^2-24x+9=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (-24)^2-4.16.9 & &\\ & = 576-576 & & \\ & = 0 & & \\ x & = \frac{-b\pm \sqrt{D}}{2.a} & & \\ & = \frac{-(-24)}{2.16} & & \\ & = \frac{3}{4} & & \\V &= \Big\{ \frac{3}{4} \Big\} & &\end{align} \\ -----------------\)
12. \(\frac{1}{24}x^2-\frac{1}{2}x+\frac{3}{2}=0\\ \Leftrightarrow \color{red}{24.} \left(\frac{1}{24}x^2-\frac{1}{2}x+\frac{3}{2}\right)=0 \color{red}{.24} \\ \Leftrightarrow 4x^2-48x+144=0 \\\text{We zoeken de oplossingen van } \color{blue}{4x^2-48x+144=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (-48)^2-4.4.144 & &\\ & = 2304-2304 & & \\ & = 0 & & \\ x & = \frac{-b\pm \sqrt{D}}{2.a} & & \\ & = \frac{-(-48)}{2.4} & & \\ & = 6 & & \\V &= \Big\{ 6 \Big\} & &\end{align} \\ -----------------\)