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Gebruik de discriminant om volgende vierkantsvergelijkingen op te lossen

1. \(18x^2-(10x-36)=2x(x-29)\)
2. \(\frac{9}{40}x^2+\frac{3}{5}x+\frac{2}{5}=0\)
3. \((-4x+1)(-5x+5)-x(18x-20)=23\)
4. \(x(4x+13)=-4(x+1)\)
5. \(-(4-23x)=-4x^2-(13-11x)\)
6. \(x^2+\frac{5}{12}x-\frac{1}{4}=0\)
7. \(9x^2-(2x+2)=x(x-17)\)
8. \(10x^2-(19x-36)=x(x+17)\)
9. \((-x+4)(-3x+2)-x(2x-6)=-19\)
10. \(x(4x+4)=(x+1)\)
11. \(\frac{9}{4}x^2+x+\frac{1}{4}=0\)
12. \(\frac{1}{9}x^2+\frac{2}{3}x+1=0\)

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Gebruik de discriminant om volgende vierkantsvergelijkingen op te lossen

Verbetersleutel

1. \(18x^2-(10x-36)=2x(x-29) \\ \Leftrightarrow 18x^2-10x+36=2x^2-58x \\ \Leftrightarrow 16x^2+48x+36=0 \\\text{We zoeken de oplossingen van } \color{blue}{16x^2+48x+36=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (48)^2-4.16.36 & &\\ & = 2304-2304 & & \\ & = 0 & & \\ x & = \frac{-b\pm \sqrt{D}}{2.a} & & \\ & = \frac{-48}{2.16} & & \\ & = -\frac{3}{2} & & \\V &= \Big\{ -\frac{3}{2} \Big\} & &\end{align} \\ -----------------\)
2. \(\frac{9}{40}x^2+\frac{3}{5}x+\frac{2}{5}=0\\ \Leftrightarrow \color{red}{40.} \left(\frac{9}{40}x^2+\frac{3}{5}x+\frac{2}{5}\right)=0 \color{red}{.40} \\ \Leftrightarrow 9x^2+24x+16=0 \\\text{We zoeken de oplossingen van } \color{blue}{9x^2+24x+16=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (24)^2-4.9.16 & &\\ & = 576-576 & & \\ & = 0 & & \\ x & = \frac{-b\pm \sqrt{D}}{2.a} & & \\ & = \frac{-24}{2.9} & & \\ & = -\frac{4}{3} & & \\V &= \Big\{ -\frac{4}{3} \Big\} & &\end{align} \\ -----------------\)
3. \((-4x+1)(-5x+5)-x(18x-20)=23\\ \Leftrightarrow 20x^2-20x-5x+5 -18x^2+20x-23=0 \\ \Leftrightarrow 2x^2+5x-18=0 \\\text{We zoeken de oplossingen van } \color{blue}{2x^2+5x-18=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (5)^2-4.2.(-18) & &\\ & = 25+144 & & \\ & = 169 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-5-\sqrt169}{2.2} & & = \frac{-5+\sqrt169}{2.2} \\ & = \frac{-18}{4} & & = \frac{8}{4} \\ & = \frac{-9}{2} & & = 2 \\ \\ V &= \Big\{ \frac{-9}{2} ; 2 \Big\} & &\end{align} \\ -----------------\)
4. \(x(4x+13)=-4(x+1) \\ \Leftrightarrow 4x^2+13x=-4x-4 \\ \Leftrightarrow 4x^2+17x+4=0 \\\text{We zoeken de oplossingen van } \color{blue}{4x^2+17x+4=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (17)^2-4.4.4 & &\\ & = 289-64 & & \\ & = 225 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-17-\sqrt225}{2.4} & & = \frac{-17+\sqrt225}{2.4} \\ & = \frac{-32}{8} & & = \frac{-2}{8} \\ & = -4 & & = \frac{-1}{4} \\ \\ V &= \Big\{ -4 ; \frac{-1}{4} \Big\} & &\end{align} \\ -----------------\)
5. \(-(4-23x)=-4x^2-(13-11x) \\ \Leftrightarrow -4+23x=-4x^2-13+11x \\ \Leftrightarrow 4x^2+12x+9=0 \\\text{We zoeken de oplossingen van } \color{blue}{4x^2+12x+9=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (12)^2-4.4.9 & &\\ & = 144-144 & & \\ & = 0 & & \\ x & = \frac{-b\pm \sqrt{D}}{2.a} & & \\ & = \frac{-12}{2.4} & & \\ & = -\frac{3}{2} & & \\V &= \Big\{ -\frac{3}{2} \Big\} & &\end{align} \\ -----------------\)
6. \(x^2+\frac{5}{12}x-\frac{1}{4}=0\\ \Leftrightarrow \color{red}{12.} \left(x^2+\frac{5}{12}x-\frac{1}{4}\right)=0 \color{red}{.12} \\ \Leftrightarrow 12x^2+5x-3=0 \\\text{We zoeken de oplossingen van } \color{blue}{12x^2+5x-3=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (5)^2-4.12.(-3) & &\\ & = 25+144 & & \\ & = 169 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-5-\sqrt169}{2.12} & & = \frac{-5+\sqrt169}{2.12} \\ & = \frac{-18}{24} & & = \frac{8}{24} \\ & = \frac{-3}{4} & & = \frac{1}{3} \\ \\ V &= \Big\{ \frac{-3}{4} ; \frac{1}{3} \Big\} & &\end{align} \\ -----------------\)
7. \(9x^2-(2x+2)=x(x-17) \\ \Leftrightarrow 9x^2-2x-2=x^2-17x \\ \Leftrightarrow 8x^2+15x-2=0 \\\text{We zoeken de oplossingen van } \color{blue}{8x^2+15x-2=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (15)^2-4.8.(-2) & &\\ & = 225+64 & & \\ & = 289 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-15-\sqrt289}{2.8} & & = \frac{-15+\sqrt289}{2.8} \\ & = \frac{-32}{16} & & = \frac{2}{16} \\ & = -2 & & = \frac{1}{8} \\ \\ V &= \Big\{ -2 ; \frac{1}{8} \Big\} & &\end{align} \\ -----------------\)
8. \(10x^2-(19x-36)=x(x+17) \\ \Leftrightarrow 10x^2-19x+36=x^2+17x \\ \Leftrightarrow 9x^2-36x+36=0 \\\text{We zoeken de oplossingen van } \color{blue}{9x^2-36x+36=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (-36)^2-4.9.36 & &\\ & = 1296-1296 & & \\ & = 0 & & \\ x & = \frac{-b\pm \sqrt{D}}{2.a} & & \\ & = \frac{-(-36)}{2.9} & & \\ & = 2 & & \\V &= \Big\{ 2 \Big\} & &\end{align} \\ -----------------\)
9. \((-x+4)(-3x+2)-x(2x-6)=-19\\ \Leftrightarrow 3x^2-2x-12x+8 -2x^2+6x+19=0 \\ \Leftrightarrow x^2+12x+27=0 \\\text{We zoeken de oplossingen van } \color{blue}{x^2+12x+27=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (12)^2-4.1.27 & &\\ & = 144-108 & & \\ & = 36 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-12-\sqrt36}{2.1} & & = \frac{-12+\sqrt36}{2.1} \\ & = \frac{-18}{2} & & = \frac{-6}{2} \\ & = -9 & & = -3 \\ \\ V &= \Big\{ -9 ; -3 \Big\} & &\end{align} \\ -----------------\)
10. \(x(4x+4)=(x+1) \\ \Leftrightarrow 4x^2+4x=x+1 \\ \Leftrightarrow 4x^2+3x-1=0 \\\text{We zoeken de oplossingen van } \color{blue}{4x^2+3x-1=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (3)^2-4.4.(-1) & &\\ & = 9+16 & & \\ & = 25 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-3-\sqrt25}{2.4} & & = \frac{-3+\sqrt25}{2.4} \\ & = \frac{-8}{8} & & = \frac{2}{8} \\ & = -1 & & = \frac{1}{4} \\ \\ V &= \Big\{ -1 ; \frac{1}{4} \Big\} & &\end{align} \\ -----------------\)
11. \(\frac{9}{4}x^2+x+\frac{1}{4}=0\\ \Leftrightarrow \color{red}{4.} \left(\frac{9}{4}x^2+x+\frac{1}{4}\right)=0 \color{red}{.4} \\ \Leftrightarrow 9x^2+4x+1=0 \\\text{We zoeken de oplossingen van } \color{blue}{9x^2+4x+1=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (4)^2-4.9.1 & &\\ & = 16-36 & & \\ & = -20 & & \\ & < 0 \\V &= \varnothing \end{align} \\ -----------------\)
12. \(\frac{1}{9}x^2+\frac{2}{3}x+1=0\\ \Leftrightarrow \color{red}{9.} \left(\frac{1}{9}x^2+\frac{2}{3}x+1\right)=0 \color{red}{.9} \\ \Leftrightarrow x^2+6x+9=0 \\\text{We zoeken de oplossingen van } \color{blue}{x^2+6x+9=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (6)^2-4.1.9 & &\\ & = 36-36 & & \\ & = 0 & & \\ x & = \frac{-b\pm \sqrt{D}}{2.a} & & \\ & = \frac{-6}{2.1} & & \\ & = -3 & & \\V &= \Big\{ -3 \Big\} & &\end{align} \\ -----------------\)