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Gebruik de discriminant om volgende vierkantsvergelijkingen op te lossen

1. \(2x^2-(4x+12)=x(x-15)\)
2. \(-(10-18x)=-x^2-(-2-17x)\)
3. \(x(x+1)=12(x+1)\)
4. \(\frac{1}{2}x^2-\frac{5}{2}x-12=0\)
5. \(x(12x+10)=-3(x+1)\)
6. \(\frac{1}{20}x^2+\frac{1}{5}x+\frac{1}{5}=0\)
7. \(-(7-31x)=-x^2-(151-7x)\)
8. \(\frac{25}{4}x=-\frac{1}{2}x^2-18\)
9. \(-\frac{1}{4}x=-\frac{1}{4}x^2-\frac{1}{16}\)
10. \(\frac{9}{2}x^2+\frac{5}{2}x-2=0\)
11. \(\frac{3}{7}x^2+2x+\frac{7}{3}=0\)
12. \(x^2+\frac{5}{4}x-\frac{9}{4}=0\)

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Gebruik de discriminant om volgende vierkantsvergelijkingen op te lossen

Verbetersleutel

1. \(2x^2-(4x+12)=x(x-15) \\ \Leftrightarrow 2x^2-4x-12=x^2-15x \\ \Leftrightarrow x^2+11x-12=0 \\\text{We zoeken de oplossingen van } \color{blue}{x^2+11x-12=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (11)^2-4.1.(-12) & &\\ & = 121+48 & & \\ & = 169 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-11-\sqrt169}{2.1} & & = \frac{-11+\sqrt169}{2.1} \\ & = \frac{-24}{2} & & = \frac{2}{2} \\ & = -12 & & = 1 \\ \\ V &= \Big\{ -12 ; 1 \Big\} & &\end{align} \\ -----------------\)
2. \(-(10-18x)=-x^2-(-2-17x) \\ \Leftrightarrow -10+18x=-x^2+2+17x \\ \Leftrightarrow x^2+x-12=0 \\\text{We zoeken de oplossingen van } \color{blue}{x^2+x-12=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (1)^2-4.1.(-12) & &\\ & = 1+48 & & \\ & = 49 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-1-\sqrt49}{2.1} & & = \frac{-1+\sqrt49}{2.1} \\ & = \frac{-8}{2} & & = \frac{6}{2} \\ & = -4 & & = 3 \\ \\ V &= \Big\{ -4 ; 3 \Big\} & &\end{align} \\ -----------------\)
3. \(x(x+1)=12(x+1) \\ \Leftrightarrow x^2+x=12x+12 \\ \Leftrightarrow x^2-11x-12=0 \\\text{We zoeken de oplossingen van } \color{blue}{x^2-11x-12=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (-11)^2-4.1.(-12) & &\\ & = 121+48 & & \\ & = 169 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-(-11)-\sqrt169}{2.1} & & = \frac{-(-11)+\sqrt169}{2.1} \\ & = \frac{-2}{2} & & = \frac{24}{2} \\ & = -1 & & = 12 \\ \\ V &= \Big\{ -1 ; 12 \Big\} & &\end{align} \\ -----------------\)
4. \(\frac{1}{2}x^2-\frac{5}{2}x-12=0\\ \Leftrightarrow \color{red}{2.} \left(\frac{1}{2}x^2-\frac{5}{2}x-12\right)=0 \color{red}{.2} \\ \Leftrightarrow x^2-5x-24=0 \\\text{We zoeken de oplossingen van } \color{blue}{x^2-5x-24=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (-5)^2-4.1.(-24) & &\\ & = 25+96 & & \\ & = 121 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-(-5)-\sqrt121}{2.1} & & = \frac{-(-5)+\sqrt121}{2.1} \\ & = \frac{-6}{2} & & = \frac{16}{2} \\ & = -3 & & = 8 \\ \\ V &= \Big\{ -3 ; 8 \Big\} & &\end{align} \\ -----------------\)
5. \(x(12x+10)=-3(x+1) \\ \Leftrightarrow 12x^2+10x=-3x-3 \\ \Leftrightarrow 12x^2+13x+3=0 \\\text{We zoeken de oplossingen van } \color{blue}{12x^2+13x+3=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (13)^2-4.12.3 & &\\ & = 169-144 & & \\ & = 25 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-13-\sqrt25}{2.12} & & = \frac{-13+\sqrt25}{2.12} \\ & = \frac{-18}{24} & & = \frac{-8}{24} \\ & = \frac{-3}{4} & & = \frac{-1}{3} \\ \\ V &= \Big\{ \frac{-3}{4} ; \frac{-1}{3} \Big\} & &\end{align} \\ -----------------\)
6. \(\frac{1}{20}x^2+\frac{1}{5}x+\frac{1}{5}=0\\ \Leftrightarrow \color{red}{20.} \left(\frac{1}{20}x^2+\frac{1}{5}x+\frac{1}{5}\right)=0 \color{red}{.20} \\ \Leftrightarrow x^2+4x+4=0 \\\text{We zoeken de oplossingen van } \color{blue}{x^2+4x+4=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (4)^2-4.1.4 & &\\ & = 16-16 & & \\ & = 0 & & \\ x & = \frac{-b\pm \sqrt{D}}{2.a} & & \\ & = \frac{-4}{2.1} & & \\ & = -2 & & \\V &= \Big\{ -2 \Big\} & &\end{align} \\ -----------------\)
7. \(-(7-31x)=-x^2-(151-7x) \\ \Leftrightarrow -7+31x=-x^2-151+7x \\ \Leftrightarrow x^2+24x+144=0 \\\text{We zoeken de oplossingen van } \color{blue}{x^2+24x+144=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (24)^2-4.1.144 & &\\ & = 576-576 & & \\ & = 0 & & \\ x & = \frac{-b\pm \sqrt{D}}{2.a} & & \\ & = \frac{-24}{2.1} & & \\ & = -12 & & \\V &= \Big\{ -12 \Big\} & &\end{align} \\ -----------------\)
8. \(\frac{25}{4}x=-\frac{1}{2}x^2-18 \\ \Leftrightarrow \frac{1}{2}x^2+\frac{25}{4}x+18=0 \\ \Leftrightarrow \color{red}{4.} \left(\frac{1}{2}x^2+\frac{25}{4}x+18\right)=0 \color{red}{.4} \\ \Leftrightarrow 2x^2+25x+72=0 \\\text{We zoeken de oplossingen van } \color{blue}{2x^2+25x+72=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (25)^2-4.2.72 & &\\ & = 625-576 & & \\ & = 49 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-25-\sqrt49}{2.2} & & = \frac{-25+\sqrt49}{2.2} \\ & = \frac{-32}{4} & & = \frac{-18}{4} \\ & = -8 & & = \frac{-9}{2} \\ \\ V &= \Big\{ -8 ; \frac{-9}{2} \Big\} & &\end{align} \\ -----------------\)
9. \(-\frac{1}{4}x=-\frac{1}{4}x^2-\frac{1}{16} \\ \Leftrightarrow \frac{1}{4}x^2-\frac{1}{4}x+\frac{1}{16}=0 \\ \Leftrightarrow \color{red}{16.} \left(\frac{1}{4}x^2-\frac{1}{4}x+\frac{1}{16}\right)=0 \color{red}{.16} \\ \Leftrightarrow 16x^2-16x+4=0 \\\text{We zoeken de oplossingen van } \color{blue}{16x^2-16x+4=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (-16)^2-4.16.4 & &\\ & = 256-256 & & \\ & = 0 & & \\ x & = \frac{-b\pm \sqrt{D}}{2.a} & & \\ & = \frac{-(-16)}{2.16} & & \\ & = \frac{1}{2} & & \\V &= \Big\{ \frac{1}{2} \Big\} & &\end{align} \\ -----------------\)
10. \(\frac{9}{2}x^2+\frac{5}{2}x-2=0\\ \Leftrightarrow \color{red}{2.} \left(\frac{9}{2}x^2+\frac{5}{2}x-2\right)=0 \color{red}{.2} \\ \Leftrightarrow 9x^2+5x-4=0 \\\text{We zoeken de oplossingen van } \color{blue}{9x^2+5x-4=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (5)^2-4.9.(-4) & &\\ & = 25+144 & & \\ & = 169 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-5-\sqrt169}{2.9} & & = \frac{-5+\sqrt169}{2.9} \\ & = \frac{-18}{18} & & = \frac{8}{18} \\ & = -1 & & = \frac{4}{9} \\ \\ V &= \Big\{ -1 ; \frac{4}{9} \Big\} & &\end{align} \\ -----------------\)
11. \(\frac{3}{7}x^2+2x+\frac{7}{3}=0\\ \Leftrightarrow \color{red}{21.} \left(\frac{3}{7}x^2+2x+\frac{7}{3}\right)=0 \color{red}{.21} \\ \Leftrightarrow 9x^2+42x+49=0 \\\text{We zoeken de oplossingen van } \color{blue}{9x^2+42x+49=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (42)^2-4.9.49 & &\\ & = 1764-1764 & & \\ & = 0 & & \\ x & = \frac{-b\pm \sqrt{D}}{2.a} & & \\ & = \frac{-42}{2.9} & & \\ & = -\frac{7}{3} & & \\V &= \Big\{ -\frac{7}{3} \Big\} & &\end{align} \\ -----------------\)
12. \(x^2+\frac{5}{4}x-\frac{9}{4}=0\\ \Leftrightarrow \color{red}{4.} \left(x^2+\frac{5}{4}x-\frac{9}{4}\right)=0 \color{red}{.4} \\ \Leftrightarrow 4x^2+5x-9=0 \\\text{We zoeken de oplossingen van } \color{blue}{4x^2+5x-9=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (5)^2-4.4.(-9) & &\\ & = 25+144 & & \\ & = 169 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-5-\sqrt169}{2.4} & & = \frac{-5+\sqrt169}{2.4} \\ & = \frac{-18}{8} & & = \frac{8}{8} \\ & = \frac{-9}{4} & & = 1 \\ \\ V &= \Big\{ \frac{-9}{4} ; 1 \Big\} & &\end{align} \\ -----------------\)