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Gebruik de discriminant om volgende vierkantsvergelijkingen op te lossen

1. \(2x^2-(11x+24)=x(x-9)\)
2. \(-(3-9x)=-x^2-(1-8x)\)
3. \(x(4x+44)=4(x-25)\)
4. \(2x^2-(10x-40)=x(x+3)\)
5. \(10x^2-(16x-144)=x(x+56)\)
6. \(x(x-9)=-4(x+1)\)
7. \(-(5-22x)=-18x^2-(3-17x)\)
8. \(-\frac{1}{2}x=-\frac{1}{10}x^2+5\)
9. \((3x+1)(-2x+5)-x(-18x+13)=17\)
10. \(2x=-\frac{1}{2}x^2+\frac{77}{2}\)
11. \(5x^2-(14x+9)=x(x-19)\)
12. \(\frac{1}{5}x^2-x+\frac{5}{4}=0\)

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Gebruik de discriminant om volgende vierkantsvergelijkingen op te lossen

Verbetersleutel

1. \(2x^2-(11x+24)=x(x-9) \\ \Leftrightarrow 2x^2-11x-24=x^2-9x \\ \Leftrightarrow x^2-2x-24=0 \\\text{We zoeken de oplossingen van } \color{blue}{x^2-2x-24=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (-2)^2-4.1.(-24) & &\\ & = 4+96 & & \\ & = 100 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-(-2)-\sqrt100}{2.1} & & = \frac{-(-2)+\sqrt100}{2.1} \\ & = \frac{-8}{2} & & = \frac{12}{2} \\ & = -4 & & = 6 \\ \\ V &= \Big\{ -4 ; 6 \Big\} & &\end{align} \\ -----------------\)
2. \(-(3-9x)=-x^2-(1-8x) \\ \Leftrightarrow -3+9x=-x^2-1+8x \\ \Leftrightarrow x^2+x-2=0 \\\text{We zoeken de oplossingen van } \color{blue}{x^2+x-2=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (1)^2-4.1.(-2) & &\\ & = 1+8 & & \\ & = 9 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-1-\sqrt9}{2.1} & & = \frac{-1+\sqrt9}{2.1} \\ & = \frac{-4}{2} & & = \frac{2}{2} \\ & = -2 & & = 1 \\ \\ V &= \Big\{ -2 ; 1 \Big\} & &\end{align} \\ -----------------\)
3. \(x(4x+44)=4(x-25) \\ \Leftrightarrow 4x^2+44x=4x-100 \\ \Leftrightarrow 4x^2+40x+100=0 \\\text{We zoeken de oplossingen van } \color{blue}{4x^2+40x+100=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (40)^2-4.4.100 & &\\ & = 1600-1600 & & \\ & = 0 & & \\ x & = \frac{-b\pm \sqrt{D}}{2.a} & & \\ & = \frac{-40}{2.4} & & \\ & = -5 & & \\V &= \Big\{ -5 \Big\} & &\end{align} \\ -----------------\)
4. \(2x^2-(10x-40)=x(x+3) \\ \Leftrightarrow 2x^2-10x+40=x^2+3x \\ \Leftrightarrow x^2-13x+40=0 \\\text{We zoeken de oplossingen van } \color{blue}{x^2-13x+40=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (-13)^2-4.1.40 & &\\ & = 169-160 & & \\ & = 9 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-(-13)-\sqrt9}{2.1} & & = \frac{-(-13)+\sqrt9}{2.1} \\ & = \frac{10}{2} & & = \frac{16}{2} \\ & = 5 & & = 8 \\ \\ V &= \Big\{ 5 ; 8 \Big\} & &\end{align} \\ -----------------\)
5. \(10x^2-(16x-144)=x(x+56) \\ \Leftrightarrow 10x^2-16x+144=x^2+56x \\ \Leftrightarrow 9x^2-72x+144=0 \\\text{We zoeken de oplossingen van } \color{blue}{9x^2-72x+144=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (-72)^2-4.9.144 & &\\ & = 5184-5184 & & \\ & = 0 & & \\ x & = \frac{-b\pm \sqrt{D}}{2.a} & & \\ & = \frac{-(-72)}{2.9} & & \\ & = 4 & & \\V &= \Big\{ 4 \Big\} & &\end{align} \\ -----------------\)
6. \(x(x-9)=-4(x+1) \\ \Leftrightarrow x^2-9x=-4x-4 \\ \Leftrightarrow x^2-5x+4=0 \\\text{We zoeken de oplossingen van } \color{blue}{x^2-5x+4=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (-5)^2-4.1.4 & &\\ & = 25-16 & & \\ & = 9 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-(-5)-\sqrt9}{2.1} & & = \frac{-(-5)+\sqrt9}{2.1} \\ & = \frac{2}{2} & & = \frac{8}{2} \\ & = 1 & & = 4 \\ \\ V &= \Big\{ 1 ; 4 \Big\} & &\end{align} \\ -----------------\)
7. \(-(5-22x)=-18x^2-(3-17x) \\ \Leftrightarrow -5+22x=-18x^2-3+17x \\ \Leftrightarrow 18x^2+5x-2=0 \\\text{We zoeken de oplossingen van } \color{blue}{18x^2+5x-2=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (5)^2-4.18.(-2) & &\\ & = 25+144 & & \\ & = 169 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-5-\sqrt169}{2.18} & & = \frac{-5+\sqrt169}{2.18} \\ & = \frac{-18}{36} & & = \frac{8}{36} \\ & = \frac{-1}{2} & & = \frac{2}{9} \\ \\ V &= \Big\{ \frac{-1}{2} ; \frac{2}{9} \Big\} & &\end{align} \\ -----------------\)
8. \(-\frac{1}{2}x=-\frac{1}{10}x^2+5 \\ \Leftrightarrow \frac{1}{10}x^2-\frac{1}{2}x-5=0 \\ \Leftrightarrow \color{red}{10.} \left(\frac{1}{10}x^2-\frac{1}{2}x-5\right)=0 \color{red}{.10} \\ \Leftrightarrow x^2-5x-50=0 \\\text{We zoeken de oplossingen van } \color{blue}{x^2-5x-50=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (-5)^2-4.1.(-50) & &\\ & = 25+200 & & \\ & = 225 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-(-5)-\sqrt225}{2.1} & & = \frac{-(-5)+\sqrt225}{2.1} \\ & = \frac{-10}{2} & & = \frac{20}{2} \\ & = -5 & & = 10 \\ \\ V &= \Big\{ -5 ; 10 \Big\} & &\end{align} \\ -----------------\)
9. \((3x+1)(-2x+5)-x(-18x+13)=17\\ \Leftrightarrow -6x^2+15x-2x+5 +18x^2-13x-17=0 \\ \Leftrightarrow 12x^2+7x-12=0 \\\text{We zoeken de oplossingen van } \color{blue}{12x^2+7x-12=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (7)^2-4.12.(-12) & &\\ & = 49+576 & & \\ & = 625 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-7-\sqrt625}{2.12} & & = \frac{-7+\sqrt625}{2.12} \\ & = \frac{-32}{24} & & = \frac{18}{24} \\ & = \frac{-4}{3} & & = \frac{3}{4} \\ \\ V &= \Big\{ \frac{-4}{3} ; \frac{3}{4} \Big\} & &\end{align} \\ -----------------\)
10. \(2x=-\frac{1}{2}x^2+\frac{77}{2} \\ \Leftrightarrow \frac{1}{2}x^2+2x-\frac{77}{2}=0 \\ \Leftrightarrow \color{red}{2.} \left(\frac{1}{2}x^2+2x-\frac{77}{2}\right)=0 \color{red}{.2} \\ \Leftrightarrow x^2+4x-77=0 \\\text{We zoeken de oplossingen van } \color{blue}{x^2+4x-77=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (4)^2-4.1.(-77) & &\\ & = 16+308 & & \\ & = 324 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-4-\sqrt324}{2.1} & & = \frac{-4+\sqrt324}{2.1} \\ & = \frac{-22}{2} & & = \frac{14}{2} \\ & = -11 & & = 7 \\ \\ V &= \Big\{ -11 ; 7 \Big\} & &\end{align} \\ -----------------\)
11. \(5x^2-(14x+9)=x(x-19) \\ \Leftrightarrow 5x^2-14x-9=x^2-19x \\ \Leftrightarrow 4x^2+5x-9=0 \\\text{We zoeken de oplossingen van } \color{blue}{4x^2+5x-9=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (5)^2-4.4.(-9) & &\\ & = 25+144 & & \\ & = 169 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-5-\sqrt169}{2.4} & & = \frac{-5+\sqrt169}{2.4} \\ & = \frac{-18}{8} & & = \frac{8}{8} \\ & = \frac{-9}{4} & & = 1 \\ \\ V &= \Big\{ \frac{-9}{4} ; 1 \Big\} & &\end{align} \\ -----------------\)
12. \(\frac{1}{5}x^2-x+\frac{5}{4}=0\\ \Leftrightarrow \color{red}{20.} \left(\frac{1}{5}x^2-x+\frac{5}{4}\right)=0 \color{red}{.20} \\ \Leftrightarrow 16x^2-80x+100=0 \\\text{We zoeken de oplossingen van } \color{blue}{16x^2-80x+100=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (-80)^2-4.16.100 & &\\ & = 6400-6400 & & \\ & = 0 & & \\ x & = \frac{-b\pm \sqrt{D}}{2.a} & & \\ & = \frac{-(-80)}{2.16} & & \\ & = \frac{5}{2} & & \\V &= \Big\{ \frac{5}{2} \Big\} & &\end{align} \\ -----------------\)