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Gebruik de discriminant om volgende vierkantsvergelijkingen op te lossen

1. \(\frac{1}{5}x^2-\frac{7}{5}x-12=0\)
2. \(\frac{1}{3}x^2-\frac{4}{3}x-\frac{5}{3}=0\)
3. \(49x^2-(8x-3)=x(x-33)\)
4. \(x^2-\frac{7}{2}x+\frac{49}{16}=0\)
5. \(\frac{1}{2}x=-\frac{1}{4}x^2+\frac{3}{4}\)
6. \(\frac{1}{8}x^2+\frac{1}{2}x+\frac{1}{2}=0\)
7. \(\frac{25}{12}x=-3x^2-\frac{1}{3}\)
8. \(\frac{1}{2}x^2+\frac{5}{2}x-33=0\)
9. \(13x^2-(15x-9)=9x(x-3)\)
10. \(8x^2-(19x-4)=7x(x-3)\)
11. \(-(13-x)=-9x^2-(113-9x)\)
12. \(7x^2-(6x-6)=x(x-19)\)

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Gebruik de discriminant om volgende vierkantsvergelijkingen op te lossen

Verbetersleutel

1. \(\frac{1}{5}x^2-\frac{7}{5}x-12=0\\ \Leftrightarrow \color{red}{5.} \left(\frac{1}{5}x^2-\frac{7}{5}x-12\right)=0 \color{red}{.5} \\ \Leftrightarrow x^2-7x-60=0 \\\text{We zoeken de oplossingen van } \color{blue}{x^2-7x-60=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (-7)^2-4.1.(-60) & &\\ & = 49+240 & & \\ & = 289 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-(-7)-\sqrt289}{2.1} & & = \frac{-(-7)+\sqrt289}{2.1} \\ & = \frac{-10}{2} & & = \frac{24}{2} \\ & = -5 & & = 12 \\ \\ V &= \Big\{ -5 ; 12 \Big\} & &\end{align} \\ -----------------\)
2. \(\frac{1}{3}x^2-\frac{4}{3}x-\frac{5}{3}=0\\ \Leftrightarrow \color{red}{3.} \left(\frac{1}{3}x^2-\frac{4}{3}x-\frac{5}{3}\right)=0 \color{red}{.3} \\ \Leftrightarrow x^2-4x-5=0 \\\text{We zoeken de oplossingen van } \color{blue}{x^2-4x-5=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (-4)^2-4.1.(-5) & &\\ & = 16+20 & & \\ & = 36 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-(-4)-\sqrt36}{2.1} & & = \frac{-(-4)+\sqrt36}{2.1} \\ & = \frac{-2}{2} & & = \frac{10}{2} \\ & = -1 & & = 5 \\ \\ V &= \Big\{ -1 ; 5 \Big\} & &\end{align} \\ -----------------\)
3. \(49x^2-(8x-3)=x(x-33) \\ \Leftrightarrow 49x^2-8x+3=x^2-33x \\ \Leftrightarrow 48x^2+25x+3=0 \\\text{We zoeken de oplossingen van } \color{blue}{48x^2+25x+3=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (25)^2-4.48.3 & &\\ & = 625-576 & & \\ & = 49 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-25-\sqrt49}{2.48} & & = \frac{-25+\sqrt49}{2.48} \\ & = \frac{-32}{96} & & = \frac{-18}{96} \\ & = \frac{-1}{3} & & = \frac{-3}{16} \\ \\ V &= \Big\{ \frac{-1}{3} ; \frac{-3}{16} \Big\} & &\end{align} \\ -----------------\)
4. \(x^2-\frac{7}{2}x+\frac{49}{16}=0\\ \Leftrightarrow \color{red}{16.} \left(x^2-\frac{7}{2}x+\frac{49}{16}\right)=0 \color{red}{.16} \\ \Leftrightarrow 16x^2-56x+49=0 \\\text{We zoeken de oplossingen van } \color{blue}{16x^2-56x+49=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (-56)^2-4.16.49 & &\\ & = 3136-3136 & & \\ & = 0 & & \\ x & = \frac{-b\pm \sqrt{D}}{2.a} & & \\ & = \frac{-(-56)}{2.16} & & \\ & = \frac{7}{4} & & \\V &= \Big\{ \frac{7}{4} \Big\} & &\end{align} \\ -----------------\)
5. \(\frac{1}{2}x=-\frac{1}{4}x^2+\frac{3}{4} \\ \Leftrightarrow \frac{1}{4}x^2+\frac{1}{2}x-\frac{3}{4}=0 \\ \Leftrightarrow \color{red}{4.} \left(\frac{1}{4}x^2+\frac{1}{2}x-\frac{3}{4}\right)=0 \color{red}{.4} \\ \Leftrightarrow x^2+2x-3=0 \\\text{We zoeken de oplossingen van } \color{blue}{x^2+2x-3=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (2)^2-4.1.(-3) & &\\ & = 4+12 & & \\ & = 16 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-2-\sqrt16}{2.1} & & = \frac{-2+\sqrt16}{2.1} \\ & = \frac{-6}{2} & & = \frac{2}{2} \\ & = -3 & & = 1 \\ \\ V &= \Big\{ -3 ; 1 \Big\} & &\end{align} \\ -----------------\)
6. \(\frac{1}{8}x^2+\frac{1}{2}x+\frac{1}{2}=0\\ \Leftrightarrow \color{red}{8.} \left(\frac{1}{8}x^2+\frac{1}{2}x+\frac{1}{2}\right)=0 \color{red}{.8} \\ \Leftrightarrow 16x^2+64x+64=0 \\\text{We zoeken de oplossingen van } \color{blue}{16x^2+64x+64=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (64)^2-4.16.64 & &\\ & = 4096-4096 & & \\ & = 0 & & \\ x & = \frac{-b\pm \sqrt{D}}{2.a} & & \\ & = \frac{-64}{2.16} & & \\ & = -2 & & \\V &= \Big\{ -2 \Big\} & &\end{align} \\ -----------------\)
7. \(\frac{25}{12}x=-3x^2-\frac{1}{3} \\ \Leftrightarrow 3x^2+\frac{25}{12}x+\frac{1}{3}=0 \\ \Leftrightarrow \color{red}{12.} \left(3x^2+\frac{25}{12}x+\frac{1}{3}\right)=0 \color{red}{.12} \\ \Leftrightarrow 36x^2+25x+4=0 \\\text{We zoeken de oplossingen van } \color{blue}{36x^2+25x+4=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (25)^2-4.36.4 & &\\ & = 625-576 & & \\ & = 49 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-25-\sqrt49}{2.36} & & = \frac{-25+\sqrt49}{2.36} \\ & = \frac{-32}{72} & & = \frac{-18}{72} \\ & = \frac{-4}{9} & & = \frac{-1}{4} \\ \\ V &= \Big\{ \frac{-4}{9} ; \frac{-1}{4} \Big\} & &\end{align} \\ -----------------\)
8. \(\frac{1}{2}x^2+\frac{5}{2}x-33=0\\ \Leftrightarrow \color{red}{2.} \left(\frac{1}{2}x^2+\frac{5}{2}x-33\right)=0 \color{red}{.2} \\ \Leftrightarrow x^2+5x-66=0 \\\text{We zoeken de oplossingen van } \color{blue}{x^2+5x-66=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (5)^2-4.1.(-66) & &\\ & = 25+264 & & \\ & = 289 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-5-\sqrt289}{2.1} & & = \frac{-5+\sqrt289}{2.1} \\ & = \frac{-22}{2} & & = \frac{12}{2} \\ & = -11 & & = 6 \\ \\ V &= \Big\{ -11 ; 6 \Big\} & &\end{align} \\ -----------------\)
9. \(13x^2-(15x-9)=9x(x-3) \\ \Leftrightarrow 13x^2-15x+9=9x^2-27x \\ \Leftrightarrow 4x^2+12x+9=0 \\\text{We zoeken de oplossingen van } \color{blue}{4x^2+12x+9=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (12)^2-4.4.9 & &\\ & = 144-144 & & \\ & = 0 & & \\ x & = \frac{-b\pm \sqrt{D}}{2.a} & & \\ & = \frac{-12}{2.4} & & \\ & = -\frac{3}{2} & & \\V &= \Big\{ -\frac{3}{2} \Big\} & &\end{align} \\ -----------------\)
10. \(8x^2-(19x-4)=7x(x-3) \\ \Leftrightarrow 8x^2-19x+4=7x^2-21x \\ \Leftrightarrow x^2+2x+4=0 \\\text{We zoeken de oplossingen van } \color{blue}{x^2+2x+4=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (2)^2-4.1.4 & &\\ & = 4-16 & & \\ & = -12 & & \\ & < 0 \\V &= \varnothing \end{align} \\ -----------------\)
11. \(-(13-x)=-9x^2-(113-9x) \\ \Leftrightarrow -13+x=-9x^2-113+9x \\ \Leftrightarrow 9x^2-8x+100=0 \\\text{We zoeken de oplossingen van } \color{blue}{9x^2-8x+100=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (-8)^2-4.9.100 & &\\ & = 64-3600 & & \\ & = -3536 & & \\ & < 0 \\V &= \varnothing \end{align} \\ -----------------\)
12. \(7x^2-(6x-6)=x(x-19) \\ \Leftrightarrow 7x^2-6x+6=x^2-19x \\ \Leftrightarrow 6x^2+13x+6=0 \\\text{We zoeken de oplossingen van } \color{blue}{6x^2+13x+6=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (13)^2-4.6.6 & &\\ & = 169-144 & & \\ & = 25 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-13-\sqrt25}{2.6} & & = \frac{-13+\sqrt25}{2.6} \\ & = \frac{-18}{12} & & = \frac{-8}{12} \\ & = \frac{-3}{2} & & = \frac{-2}{3} \\ \\ V &= \Big\{ \frac{-3}{2} ; \frac{-2}{3} \Big\} & &\end{align} \\ -----------------\)