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Gebruik de discriminant om volgende vierkantsvergelijkingen op te lossen

1. \(x(x+25)=28(x+1)\)
2. \(x(72x+23)=-2(x+1)\)
3. \(5x^2-(8x-49)=x(x-36)\)
4. \(2x^2-(9x-48)=x(x+5)\)
5. \((x+4)(-3x+3)-x(-15x+8)=24\)
6. \(2x^2-(14x+66)=x(x-9)\)
7. \(\frac{1}{3}x^2-2x+\frac{25}{3}=0\)
8. \(x(x-15)=-25(x+1)\)
9. \(\frac{3}{5}x^2-\frac{4}{5}x+\frac{5}{3}=0\)
10. \(2x^2-(6x-25)=x(x+2)\)
11. \(-(5+2x)=-x^2-(95-17x)\)
12. \(x(18x+11)=-2(x+1)\)

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Gebruik de discriminant om volgende vierkantsvergelijkingen op te lossen

Verbetersleutel

1. \(x(x+25)=28(x+1) \\ \Leftrightarrow x^2+25x=28x+28 \\ \Leftrightarrow x^2-3x-28=0 \\\text{We zoeken de oplossingen van } \color{blue}{x^2-3x-28=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (-3)^2-4.1.(-28) & &\\ & = 9+112 & & \\ & = 121 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-(-3)-\sqrt121}{2.1} & & = \frac{-(-3)+\sqrt121}{2.1} \\ & = \frac{-8}{2} & & = \frac{14}{2} \\ & = -4 & & = 7 \\ \\ V &= \Big\{ -4 ; 7 \Big\} & &\end{align} \\ -----------------\)
2. \(x(72x+23)=-2(x+1) \\ \Leftrightarrow 72x^2+23x=-2x-2 \\ \Leftrightarrow 72x^2+25x+2=0 \\\text{We zoeken de oplossingen van } \color{blue}{72x^2+25x+2=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (25)^2-4.72.2 & &\\ & = 625-576 & & \\ & = 49 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-25-\sqrt49}{2.72} & & = \frac{-25+\sqrt49}{2.72} \\ & = \frac{-32}{144} & & = \frac{-18}{144} \\ & = \frac{-2}{9} & & = \frac{-1}{8} \\ \\ V &= \Big\{ \frac{-2}{9} ; \frac{-1}{8} \Big\} & &\end{align} \\ -----------------\)
3. \(5x^2-(8x-49)=x(x-36) \\ \Leftrightarrow 5x^2-8x+49=x^2-36x \\ \Leftrightarrow 4x^2+28x+49=0 \\\text{We zoeken de oplossingen van } \color{blue}{4x^2+28x+49=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (28)^2-4.4.49 & &\\ & = 784-784 & & \\ & = 0 & & \\ x & = \frac{-b\pm \sqrt{D}}{2.a} & & \\ & = \frac{-28}{2.4} & & \\ & = -\frac{7}{2} & & \\V &= \Big\{ -\frac{7}{2} \Big\} & &\end{align} \\ -----------------\)
4. \(2x^2-(9x-48)=x(x+5) \\ \Leftrightarrow 2x^2-9x+48=x^2+5x \\ \Leftrightarrow x^2-14x+48=0 \\\text{We zoeken de oplossingen van } \color{blue}{x^2-14x+48=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (-14)^2-4.1.48 & &\\ & = 196-192 & & \\ & = 4 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-(-14)-\sqrt4}{2.1} & & = \frac{-(-14)+\sqrt4}{2.1} \\ & = \frac{12}{2} & & = \frac{16}{2} \\ & = 6 & & = 8 \\ \\ V &= \Big\{ 6 ; 8 \Big\} & &\end{align} \\ -----------------\)
5. \((x+4)(-3x+3)-x(-15x+8)=24\\ \Leftrightarrow -3x^2+3x-12x+12 +15x^2-8x-24=0 \\ \Leftrightarrow 12x^2+7x-12=0 \\\text{We zoeken de oplossingen van } \color{blue}{12x^2+7x-12=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (7)^2-4.12.(-12) & &\\ & = 49+576 & & \\ & = 625 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-7-\sqrt625}{2.12} & & = \frac{-7+\sqrt625}{2.12} \\ & = \frac{-32}{24} & & = \frac{18}{24} \\ & = \frac{-4}{3} & & = \frac{3}{4} \\ \\ V &= \Big\{ \frac{-4}{3} ; \frac{3}{4} \Big\} & &\end{align} \\ -----------------\)
6. \(2x^2-(14x+66)=x(x-9) \\ \Leftrightarrow 2x^2-14x-66=x^2-9x \\ \Leftrightarrow x^2-5x-66=0 \\\text{We zoeken de oplossingen van } \color{blue}{x^2-5x-66=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (-5)^2-4.1.(-66) & &\\ & = 25+264 & & \\ & = 289 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-(-5)-\sqrt289}{2.1} & & = \frac{-(-5)+\sqrt289}{2.1} \\ & = \frac{-12}{2} & & = \frac{22}{2} \\ & = -6 & & = 11 \\ \\ V &= \Big\{ -6 ; 11 \Big\} & &\end{align} \\ -----------------\)
7. \(\frac{1}{3}x^2-2x+\frac{25}{3}=0\\ \Leftrightarrow \color{red}{3.} \left(\frac{1}{3}x^2-2x+\frac{25}{3}\right)=0 \color{red}{.3} \\ \Leftrightarrow x^2-6x+25=0 \\\text{We zoeken de oplossingen van } \color{blue}{x^2-6x+25=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (-6)^2-4.1.25 & &\\ & = 36-100 & & \\ & = -64 & & \\ & < 0 \\V &= \varnothing \end{align} \\ -----------------\)
8. \(x(x-15)=-25(x+1) \\ \Leftrightarrow x^2-15x=-25x-25 \\ \Leftrightarrow x^2+10x+25=0 \\\text{We zoeken de oplossingen van } \color{blue}{x^2+10x+25=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (10)^2-4.1.25 & &\\ & = 100-100 & & \\ & = 0 & & \\ x & = \frac{-b\pm \sqrt{D}}{2.a} & & \\ & = \frac{-10}{2.1} & & \\ & = -5 & & \\V &= \Big\{ -5 \Big\} & &\end{align} \\ -----------------\)
9. \(\frac{3}{5}x^2-\frac{4}{5}x+\frac{5}{3}=0\\ \Leftrightarrow \color{red}{15.} \left(\frac{3}{5}x^2-\frac{4}{5}x+\frac{5}{3}\right)=0 \color{red}{.15} \\ \Leftrightarrow 9x^2-12x+25=0 \\\text{We zoeken de oplossingen van } \color{blue}{9x^2-12x+25=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (-12)^2-4.9.25 & &\\ & = 144-900 & & \\ & = -756 & & \\ & < 0 \\V &= \varnothing \end{align} \\ -----------------\)
10. \(2x^2-(6x-25)=x(x+2) \\ \Leftrightarrow 2x^2-6x+25=x^2+2x \\ \Leftrightarrow x^2-8x+25=0 \\\text{We zoeken de oplossingen van } \color{blue}{x^2-8x+25=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (-8)^2-4.1.25 & &\\ & = 64-100 & & \\ & = -36 & & \\ & < 0 \\V &= \varnothing \end{align} \\ -----------------\)
11. \(-(5+2x)=-x^2-(95-17x) \\ \Leftrightarrow -5-2x=-x^2-95+17x \\ \Leftrightarrow x^2-19x+90=0 \\\text{We zoeken de oplossingen van } \color{blue}{x^2-19x+90=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (-19)^2-4.1.90 & &\\ & = 361-360 & & \\ & = 1 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-(-19)-\sqrt1}{2.1} & & = \frac{-(-19)+\sqrt1}{2.1} \\ & = \frac{18}{2} & & = \frac{20}{2} \\ & = 9 & & = 10 \\ \\ V &= \Big\{ 9 ; 10 \Big\} & &\end{align} \\ -----------------\)
12. \(x(18x+11)=-2(x+1) \\ \Leftrightarrow 18x^2+11x=-2x-2 \\ \Leftrightarrow 18x^2+13x+2=0 \\\text{We zoeken de oplossingen van } \color{blue}{18x^2+13x+2=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (13)^2-4.18.2 & &\\ & = 169-144 & & \\ & = 25 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-13-\sqrt25}{2.18} & & = \frac{-13+\sqrt25}{2.18} \\ & = \frac{-18}{36} & & = \frac{-8}{36} \\ & = \frac{-1}{2} & & = \frac{-2}{9} \\ \\ V &= \Big\{ \frac{-1}{2} ; \frac{-2}{9} \Big\} & &\end{align} \\ -----------------\)