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Gebruik de discriminant om volgende vierkantsvergelijkingen op te lossen

1. \(3x^2-(16x-16)=2x(x-3)\)
2. \(\frac{1}{3}x^2-\frac{1}{3}x+\frac{1}{12}=0\)
3. \(16x^2-(17x-49)=7x(x-7)\)
4. \(\frac{1}{2}x^2-7x+\frac{33}{2}=0\)
5. \((3x+5)(-x+5)-x(-4x+58)=-56\)
6. \(\frac{1}{4}x^2+\frac{25}{12}x+4=0\)
7. \(-(13+6x)=-16x^2-(22-18x)\)
8. \(-(15-20x)=-4x^2-(24-16x)\)
9. \(\frac{1}{72}x^2-\frac{1}{3}x+2=0\)
10. \(\frac{1}{27}x^2+\frac{1}{3}x-\frac{4}{3}=0\)
11. \(\frac{1}{4}x^2+\frac{15}{8}x-1=0\)
12. \(\frac{1}{40}x^2+\frac{1}{5}x+\frac{2}{5}=0\)

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Gebruik de discriminant om volgende vierkantsvergelijkingen op te lossen

Verbetersleutel

1. \(3x^2-(16x-16)=2x(x-3) \\ \Leftrightarrow 3x^2-16x+16=2x^2-6x \\ \Leftrightarrow x^2-10x+16=0 \\\text{We zoeken de oplossingen van } \color{blue}{x^2-10x+16=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (-10)^2-4.1.16 & &\\ & = 100-64 & & \\ & = 36 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-(-10)-\sqrt36}{2.1} & & = \frac{-(-10)+\sqrt36}{2.1} \\ & = \frac{4}{2} & & = \frac{16}{2} \\ & = 2 & & = 8 \\ \\ V &= \Big\{ 2 ; 8 \Big\} & &\end{align} \\ -----------------\)
2. \(\frac{1}{3}x^2-\frac{1}{3}x+\frac{1}{12}=0\\ \Leftrightarrow \color{red}{12.} \left(\frac{1}{3}x^2-\frac{1}{3}x+\frac{1}{12}\right)=0 \color{red}{.12} \\ \Leftrightarrow 4x^2-4x+1=0 \\\text{We zoeken de oplossingen van } \color{blue}{4x^2-4x+1=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (-4)^2-4.4.1 & &\\ & = 16-16 & & \\ & = 0 & & \\ x & = \frac{-b\pm \sqrt{D}}{2.a} & & \\ & = \frac{-(-4)}{2.4} & & \\ & = \frac{1}{2} & & \\V &= \Big\{ \frac{1}{2} \Big\} & &\end{align} \\ -----------------\)
3. \(16x^2-(17x-49)=7x(x-7) \\ \Leftrightarrow 16x^2-17x+49=7x^2-49x \\ \Leftrightarrow 9x^2+32x+49=0 \\\text{We zoeken de oplossingen van } \color{blue}{9x^2+32x+49=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (32)^2-4.9.49 & &\\ & = 1024-1764 & & \\ & = -740 & & \\ & < 0 \\V &= \varnothing \end{align} \\ -----------------\)
4. \(\frac{1}{2}x^2-7x+\frac{33}{2}=0\\ \Leftrightarrow \color{red}{2.} \left(\frac{1}{2}x^2-7x+\frac{33}{2}\right)=0 \color{red}{.2} \\ \Leftrightarrow x^2-14x+33=0 \\\text{We zoeken de oplossingen van } \color{blue}{x^2-14x+33=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (-14)^2-4.1.33 & &\\ & = 196-132 & & \\ & = 64 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-(-14)-\sqrt64}{2.1} & & = \frac{-(-14)+\sqrt64}{2.1} \\ & = \frac{6}{2} & & = \frac{22}{2} \\ & = 3 & & = 11 \\ \\ V &= \Big\{ 3 ; 11 \Big\} & &\end{align} \\ -----------------\)
5. \((3x+5)(-x+5)-x(-4x+58)=-56\\ \Leftrightarrow -3x^2+15x-5x+25 +4x^2-58x+56=0 \\ \Leftrightarrow x^2-18x+81=0 \\\text{We zoeken de oplossingen van } \color{blue}{x^2-18x+81=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (-18)^2-4.1.81 & &\\ & = 324-324 & & \\ & = 0 & & \\ x & = \frac{-b\pm \sqrt{D}}{2.a} & & \\ & = \frac{-(-18)}{2.1} & & \\ & = 9 & & \\V &= \Big\{ 9 \Big\} & &\end{align} \\ -----------------\)
6. \(\frac{1}{4}x^2+\frac{25}{12}x+4=0\\ \Leftrightarrow \color{red}{12.} \left(\frac{1}{4}x^2+\frac{25}{12}x+4\right)=0 \color{red}{.12} \\ \Leftrightarrow 3x^2+25x+48=0 \\\text{We zoeken de oplossingen van } \color{blue}{3x^2+25x+48=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (25)^2-4.3.48 & &\\ & = 625-576 & & \\ & = 49 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-25-\sqrt49}{2.3} & & = \frac{-25+\sqrt49}{2.3} \\ & = \frac{-32}{6} & & = \frac{-18}{6} \\ & = \frac{-16}{3} & & = -3 \\ \\ V &= \Big\{ \frac{-16}{3} ; -3 \Big\} & &\end{align} \\ -----------------\)
7. \(-(13+6x)=-16x^2-(22-18x) \\ \Leftrightarrow -13-6x=-16x^2-22+18x \\ \Leftrightarrow 16x^2-24x+9=0 \\\text{We zoeken de oplossingen van } \color{blue}{16x^2-24x+9=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (-24)^2-4.16.9 & &\\ & = 576-576 & & \\ & = 0 & & \\ x & = \frac{-b\pm \sqrt{D}}{2.a} & & \\ & = \frac{-(-24)}{2.16} & & \\ & = \frac{3}{4} & & \\V &= \Big\{ \frac{3}{4} \Big\} & &\end{align} \\ -----------------\)
8. \(-(15-20x)=-4x^2-(24-16x) \\ \Leftrightarrow -15+20x=-4x^2-24+16x \\ \Leftrightarrow 4x^2+4x+9=0 \\\text{We zoeken de oplossingen van } \color{blue}{4x^2+4x+9=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (4)^2-4.4.9 & &\\ & = 16-144 & & \\ & = -128 & & \\ & < 0 \\V &= \varnothing \end{align} \\ -----------------\)
9. \(\frac{1}{72}x^2-\frac{1}{3}x+2=0\\ \Leftrightarrow \color{red}{72.} \left(\frac{1}{72}x^2-\frac{1}{3}x+2\right)=0 \color{red}{.72} \\ \Leftrightarrow x^2-24x+144=0 \\\text{We zoeken de oplossingen van } \color{blue}{x^2-24x+144=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (-24)^2-4.1.144 & &\\ & = 576-576 & & \\ & = 0 & & \\ x & = \frac{-b\pm \sqrt{D}}{2.a} & & \\ & = \frac{-(-24)}{2.1} & & \\ & = 12 & & \\V &= \Big\{ 12 \Big\} & &\end{align} \\ -----------------\)
10. \(\frac{1}{27}x^2+\frac{1}{3}x-\frac{4}{3}=0\\ \Leftrightarrow \color{red}{27.} \left(\frac{1}{27}x^2+\frac{1}{3}x-\frac{4}{3}\right)=0 \color{red}{.27} \\ \Leftrightarrow x^2+9x-36=0 \\\text{We zoeken de oplossingen van } \color{blue}{x^2+9x-36=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (9)^2-4.1.(-36) & &\\ & = 81+144 & & \\ & = 225 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-9-\sqrt225}{2.1} & & = \frac{-9+\sqrt225}{2.1} \\ & = \frac{-24}{2} & & = \frac{6}{2} \\ & = -12 & & = 3 \\ \\ V &= \Big\{ -12 ; 3 \Big\} & &\end{align} \\ -----------------\)
11. \(\frac{1}{4}x^2+\frac{15}{8}x-1=0\\ \Leftrightarrow \color{red}{8.} \left(\frac{1}{4}x^2+\frac{15}{8}x-1\right)=0 \color{red}{.8} \\ \Leftrightarrow 2x^2+15x-8=0 \\\text{We zoeken de oplossingen van } \color{blue}{2x^2+15x-8=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (15)^2-4.2.(-8) & &\\ & = 225+64 & & \\ & = 289 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-15-\sqrt289}{2.2} & & = \frac{-15+\sqrt289}{2.2} \\ & = \frac{-32}{4} & & = \frac{2}{4} \\ & = -8 & & = \frac{1}{2} \\ \\ V &= \Big\{ -8 ; \frac{1}{2} \Big\} & &\end{align} \\ -----------------\)
12. \(\frac{1}{40}x^2+\frac{1}{5}x+\frac{2}{5}=0\\ \Leftrightarrow \color{red}{40.} \left(\frac{1}{40}x^2+\frac{1}{5}x+\frac{2}{5}\right)=0 \color{red}{.40} \\ \Leftrightarrow 9x^2+72x+144=0 \\\text{We zoeken de oplossingen van } \color{blue}{9x^2+72x+144=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (72)^2-4.9.144 & &\\ & = 5184-5184 & & \\ & = 0 & & \\ x & = \frac{-b\pm \sqrt{D}}{2.a} & & \\ & = \frac{-72}{2.9} & & \\ & = -4 & & \\V &= \Big\{ -4 \Big\} & &\end{align} \\ -----------------\)