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Gebruik de discriminant om volgende vierkantsvergelijkingen op te lossen

1. \(9x^2-(10x-2)=x(x-27)\)
2. \(10x^2-(13x-49)=x(x-55)\)
3. \(x(16x-14)=2(x-2)\)
4. \(14x^2-(5x-36)=10x(x-3)\)
5. \(5x^2-(13x+18)=4x(x-5)\)
6. \(17x^2-(4x+1)=x(x-19)\)
7. \(\frac{5}{3}x=-\frac{4}{3}x^2-\frac{1}{3}\)
8. \(-(15+37x)=-16x^2-(51-11x)\)
9. \((x-4)(-2x+4)-x(-18x-29)=-17\)
10. \(x(x+131)=132(x+1)\)
11. \(x(x-16)=4(x-25)\)
12. \(-x=-\frac{1}{4}x^2+\frac{5}{4}\)

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Gebruik de discriminant om volgende vierkantsvergelijkingen op te lossen

Verbetersleutel

1. \(9x^2-(10x-2)=x(x-27) \\ \Leftrightarrow 9x^2-10x+2=x^2-27x \\ \Leftrightarrow 8x^2+17x+2=0 \\\text{We zoeken de oplossingen van } \color{blue}{8x^2+17x+2=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (17)^2-4.8.2 & &\\ & = 289-64 & & \\ & = 225 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-17-\sqrt225}{2.8} & & = \frac{-17+\sqrt225}{2.8} \\ & = \frac{-32}{16} & & = \frac{-2}{16} \\ & = -2 & & = \frac{-1}{8} \\ \\ V &= \Big\{ -2 ; \frac{-1}{8} \Big\} & &\end{align} \\ -----------------\)
2. \(10x^2-(13x-49)=x(x-55) \\ \Leftrightarrow 10x^2-13x+49=x^2-55x \\ \Leftrightarrow 9x^2+42x+49=0 \\\text{We zoeken de oplossingen van } \color{blue}{9x^2+42x+49=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (42)^2-4.9.49 & &\\ & = 1764-1764 & & \\ & = 0 & & \\ x & = \frac{-b\pm \sqrt{D}}{2.a} & & \\ & = \frac{-42}{2.9} & & \\ & = -\frac{7}{3} & & \\V &= \Big\{ -\frac{7}{3} \Big\} & &\end{align} \\ -----------------\)
3. \(x(16x-14)=2(x-2) \\ \Leftrightarrow 16x^2-14x=2x-4 \\ \Leftrightarrow 16x^2-16x+4=0 \\\text{We zoeken de oplossingen van } \color{blue}{16x^2-16x+4=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (-16)^2-4.16.4 & &\\ & = 256-256 & & \\ & = 0 & & \\ x & = \frac{-b\pm \sqrt{D}}{2.a} & & \\ & = \frac{-(-16)}{2.16} & & \\ & = \frac{1}{2} & & \\V &= \Big\{ \frac{1}{2} \Big\} & &\end{align} \\ -----------------\)
4. \(14x^2-(5x-36)=10x(x-3) \\ \Leftrightarrow 14x^2-5x+36=10x^2-30x \\ \Leftrightarrow 4x^2+25x+36=0 \\\text{We zoeken de oplossingen van } \color{blue}{4x^2+25x+36=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (25)^2-4.4.36 & &\\ & = 625-576 & & \\ & = 49 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-25-\sqrt49}{2.4} & & = \frac{-25+\sqrt49}{2.4} \\ & = \frac{-32}{8} & & = \frac{-18}{8} \\ & = -4 & & = \frac{-9}{4} \\ \\ V &= \Big\{ -4 ; \frac{-9}{4} \Big\} & &\end{align} \\ -----------------\)
5. \(5x^2-(13x+18)=4x(x-5) \\ \Leftrightarrow 5x^2-13x-18=4x^2-20x \\ \Leftrightarrow x^2+7x-18=0 \\\text{We zoeken de oplossingen van } \color{blue}{x^2+7x-18=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (7)^2-4.1.(-18) & &\\ & = 49+72 & & \\ & = 121 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-7-\sqrt121}{2.1} & & = \frac{-7+\sqrt121}{2.1} \\ & = \frac{-18}{2} & & = \frac{4}{2} \\ & = -9 & & = 2 \\ \\ V &= \Big\{ -9 ; 2 \Big\} & &\end{align} \\ -----------------\)
6. \(17x^2-(4x+1)=x(x-19) \\ \Leftrightarrow 17x^2-4x-1=x^2-19x \\ \Leftrightarrow 16x^2+15x-1=0 \\\text{We zoeken de oplossingen van } \color{blue}{16x^2+15x-1=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (15)^2-4.16.(-1) & &\\ & = 225+64 & & \\ & = 289 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-15-\sqrt289}{2.16} & & = \frac{-15+\sqrt289}{2.16} \\ & = \frac{-32}{32} & & = \frac{2}{32} \\ & = -1 & & = \frac{1}{16} \\ \\ V &= \Big\{ -1 ; \frac{1}{16} \Big\} & &\end{align} \\ -----------------\)
7. \(\frac{5}{3}x=-\frac{4}{3}x^2-\frac{1}{3} \\ \Leftrightarrow \frac{4}{3}x^2+\frac{5}{3}x+\frac{1}{3}=0 \\ \Leftrightarrow \color{red}{3.} \left(\frac{4}{3}x^2+\frac{5}{3}x+\frac{1}{3}\right)=0 \color{red}{.3} \\ \Leftrightarrow 4x^2+5x+1=0 \\\text{We zoeken de oplossingen van } \color{blue}{4x^2+5x+1=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (5)^2-4.4.1 & &\\ & = 25-16 & & \\ & = 9 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-5-\sqrt9}{2.4} & & = \frac{-5+\sqrt9}{2.4} \\ & = \frac{-8}{8} & & = \frac{-2}{8} \\ & = -1 & & = \frac{-1}{4} \\ \\ V &= \Big\{ -1 ; \frac{-1}{4} \Big\} & &\end{align} \\ -----------------\)
8. \(-(15+37x)=-16x^2-(51-11x) \\ \Leftrightarrow -15-37x=-16x^2-51+11x \\ \Leftrightarrow 16x^2-48x+36=0 \\\text{We zoeken de oplossingen van } \color{blue}{16x^2-48x+36=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (-48)^2-4.16.36 & &\\ & = 2304-2304 & & \\ & = 0 & & \\ x & = \frac{-b\pm \sqrt{D}}{2.a} & & \\ & = \frac{-(-48)}{2.16} & & \\ & = \frac{3}{2} & & \\V &= \Big\{ \frac{3}{2} \Big\} & &\end{align} \\ -----------------\)
9. \((x-4)(-2x+4)-x(-18x-29)=-17\\ \Leftrightarrow -2x^2+4x+8x-16 +18x^2+29x+17=0 \\ \Leftrightarrow 16x^2+17x+1=0 \\\text{We zoeken de oplossingen van } \color{blue}{16x^2+17x+1=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (17)^2-4.16.1 & &\\ & = 289-64 & & \\ & = 225 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-17-\sqrt225}{2.16} & & = \frac{-17+\sqrt225}{2.16} \\ & = \frac{-32}{32} & & = \frac{-2}{32} \\ & = -1 & & = \frac{-1}{16} \\ \\ V &= \Big\{ -1 ; \frac{-1}{16} \Big\} & &\end{align} \\ -----------------\)
10. \(x(x+131)=132(x+1) \\ \Leftrightarrow x^2+131x=132x+132 \\ \Leftrightarrow x^2-x-132=0 \\\text{We zoeken de oplossingen van } \color{blue}{x^2-x-132=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (-1)^2-4.1.(-132) & &\\ & = 1+528 & & \\ & = 529 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-(-1)-\sqrt529}{2.1} & & = \frac{-(-1)+\sqrt529}{2.1} \\ & = \frac{-22}{2} & & = \frac{24}{2} \\ & = -11 & & = 12 \\ \\ V &= \Big\{ -11 ; 12 \Big\} & &\end{align} \\ -----------------\)
11. \(x(x-16)=4(x-25) \\ \Leftrightarrow x^2-16x=4x-100 \\ \Leftrightarrow x^2-20x+100=0 \\\text{We zoeken de oplossingen van } \color{blue}{x^2-20x+100=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (-20)^2-4.1.100 & &\\ & = 400-400 & & \\ & = 0 & & \\ x & = \frac{-b\pm \sqrt{D}}{2.a} & & \\ & = \frac{-(-20)}{2.1} & & \\ & = 10 & & \\V &= \Big\{ 10 \Big\} & &\end{align} \\ -----------------\)
12. \(-x=-\frac{1}{4}x^2+\frac{5}{4} \\ \Leftrightarrow \frac{1}{4}x^2-x-\frac{5}{4}=0 \\ \Leftrightarrow \color{red}{4.} \left(\frac{1}{4}x^2-x-\frac{5}{4}\right)=0 \color{red}{.4} \\ \Leftrightarrow x^2-4x-5=0 \\\text{We zoeken de oplossingen van } \color{blue}{x^2-4x-5=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (-4)^2-4.1.(-5) & &\\ & = 16+20 & & \\ & = 36 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-(-4)-\sqrt36}{2.1} & & = \frac{-(-4)+\sqrt36}{2.1} \\ & = \frac{-2}{2} & & = \frac{10}{2} \\ & = -1 & & = 5 \\ \\ V &= \Big\{ -1 ; 5 \Big\} & &\end{align} \\ -----------------\)