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Gebruik de discriminant om volgende vierkantsvergelijkingen op te lossen

1. \(2x^2-(9x-36)=x(x+6)\)
2. \((x-3)(-4x+2)-x(-5x-16)=-26\)
3. \((4x-3)(3x-3)-x(11x-15)=-18\)
4. \(2x^2-(11x+96)=x(x-15)\)
5. \(x(16x+3)=-(x+1)\)
6. \(x(x+19)=14(x+1)\)
7. \(\frac{25}{12}x=-3x^2-\frac{1}{3}\)
8. \(-(13+8x)=-x^2-(53-6x)\)
9. \(9x^2-(2x-9)=5x(x-3)\)
10. \(-(7+3x)=-4x^2-(32-5x)\)
11. \(-(13+9x)=-x^2-(37-2x)\)
12. \((5x+2)(4x+1)-x(18x+4)=4\)

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Gebruik de discriminant om volgende vierkantsvergelijkingen op te lossen

Verbetersleutel

1. \(2x^2-(9x-36)=x(x+6) \\ \Leftrightarrow 2x^2-9x+36=x^2+6x \\ \Leftrightarrow x^2-15x+36=0 \\\text{We zoeken de oplossingen van } \color{blue}{x^2-15x+36=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (-15)^2-4.1.36 & &\\ & = 225-144 & & \\ & = 81 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-(-15)-\sqrt81}{2.1} & & = \frac{-(-15)+\sqrt81}{2.1} \\ & = \frac{6}{2} & & = \frac{24}{2} \\ & = 3 & & = 12 \\ \\ V &= \Big\{ 3 ; 12 \Big\} & &\end{align} \\ -----------------\)
2. \((x-3)(-4x+2)-x(-5x-16)=-26\\ \Leftrightarrow -4x^2+2x+12x-6 +5x^2+16x+26=0 \\ \Leftrightarrow x^2+12x+20=0 \\\text{We zoeken de oplossingen van } \color{blue}{x^2+12x+20=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (12)^2-4.1.20 & &\\ & = 144-80 & & \\ & = 64 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-12-\sqrt64}{2.1} & & = \frac{-12+\sqrt64}{2.1} \\ & = \frac{-20}{2} & & = \frac{-4}{2} \\ & = -10 & & = -2 \\ \\ V &= \Big\{ -10 ; -2 \Big\} & &\end{align} \\ -----------------\)
3. \((4x-3)(3x-3)-x(11x-15)=-18\\ \Leftrightarrow 12x^2-12x-9x+9 -11x^2+15x+18=0 \\ \Leftrightarrow x^2+12x+27=0 \\\text{We zoeken de oplossingen van } \color{blue}{x^2+12x+27=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (12)^2-4.1.27 & &\\ & = 144-108 & & \\ & = 36 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-12-\sqrt36}{2.1} & & = \frac{-12+\sqrt36}{2.1} \\ & = \frac{-18}{2} & & = \frac{-6}{2} \\ & = -9 & & = -3 \\ \\ V &= \Big\{ -9 ; -3 \Big\} & &\end{align} \\ -----------------\)
4. \(2x^2-(11x+96)=x(x-15) \\ \Leftrightarrow 2x^2-11x-96=x^2-15x \\ \Leftrightarrow x^2+4x-96=0 \\\text{We zoeken de oplossingen van } \color{blue}{x^2+4x-96=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (4)^2-4.1.(-96) & &\\ & = 16+384 & & \\ & = 400 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-4-\sqrt400}{2.1} & & = \frac{-4+\sqrt400}{2.1} \\ & = \frac{-24}{2} & & = \frac{16}{2} \\ & = -12 & & = 8 \\ \\ V &= \Big\{ -12 ; 8 \Big\} & &\end{align} \\ -----------------\)
5. \(x(16x+3)=-(x+1) \\ \Leftrightarrow 16x^2+3x=-x-1 \\ \Leftrightarrow 16x^2+4x+1=0 \\\text{We zoeken de oplossingen van } \color{blue}{16x^2+4x+1=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (4)^2-4.16.1 & &\\ & = 16-64 & & \\ & = -48 & & \\ & < 0 \\V &= \varnothing \end{align} \\ -----------------\)
6. \(x(x+19)=14(x+1) \\ \Leftrightarrow x^2+19x=14x+14 \\ \Leftrightarrow x^2+5x-14=0 \\\text{We zoeken de oplossingen van } \color{blue}{x^2+5x-14=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (5)^2-4.1.(-14) & &\\ & = 25+56 & & \\ & = 81 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-5-\sqrt81}{2.1} & & = \frac{-5+\sqrt81}{2.1} \\ & = \frac{-14}{2} & & = \frac{4}{2} \\ & = -7 & & = 2 \\ \\ V &= \Big\{ -7 ; 2 \Big\} & &\end{align} \\ -----------------\)
7. \(\frac{25}{12}x=-3x^2-\frac{1}{3} \\ \Leftrightarrow 3x^2+\frac{25}{12}x+\frac{1}{3}=0 \\ \Leftrightarrow \color{red}{12.} \left(3x^2+\frac{25}{12}x+\frac{1}{3}\right)=0 \color{red}{.12} \\ \Leftrightarrow 36x^2+25x+4=0 \\\text{We zoeken de oplossingen van } \color{blue}{36x^2+25x+4=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (25)^2-4.36.4 & &\\ & = 625-576 & & \\ & = 49 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-25-\sqrt49}{2.36} & & = \frac{-25+\sqrt49}{2.36} \\ & = \frac{-32}{72} & & = \frac{-18}{72} \\ & = \frac{-4}{9} & & = \frac{-1}{4} \\ \\ V &= \Big\{ \frac{-4}{9} ; \frac{-1}{4} \Big\} & &\end{align} \\ -----------------\)
8. \(-(13+8x)=-x^2-(53-6x) \\ \Leftrightarrow -13-8x=-x^2-53+6x \\ \Leftrightarrow x^2-14x+40=0 \\\text{We zoeken de oplossingen van } \color{blue}{x^2-14x+40=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (-14)^2-4.1.40 & &\\ & = 196-160 & & \\ & = 36 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-(-14)-\sqrt36}{2.1} & & = \frac{-(-14)+\sqrt36}{2.1} \\ & = \frac{8}{2} & & = \frac{20}{2} \\ & = 4 & & = 10 \\ \\ V &= \Big\{ 4 ; 10 \Big\} & &\end{align} \\ -----------------\)
9. \(9x^2-(2x-9)=5x(x-3) \\ \Leftrightarrow 9x^2-2x+9=5x^2-15x \\ \Leftrightarrow 4x^2+13x+9=0 \\\text{We zoeken de oplossingen van } \color{blue}{4x^2+13x+9=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (13)^2-4.4.9 & &\\ & = 169-144 & & \\ & = 25 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-13-\sqrt25}{2.4} & & = \frac{-13+\sqrt25}{2.4} \\ & = \frac{-18}{8} & & = \frac{-8}{8} \\ & = \frac{-9}{4} & & = -1 \\ \\ V &= \Big\{ \frac{-9}{4} ; -1 \Big\} & &\end{align} \\ -----------------\)
10. \(-(7+3x)=-4x^2-(32-5x) \\ \Leftrightarrow -7-3x=-4x^2-32+5x \\ \Leftrightarrow 4x^2-8x+25=0 \\\text{We zoeken de oplossingen van } \color{blue}{4x^2-8x+25=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (-8)^2-4.4.25 & &\\ & = 64-400 & & \\ & = -336 & & \\ & < 0 \\V &= \varnothing \end{align} \\ -----------------\)
11. \(-(13+9x)=-x^2-(37-2x) \\ \Leftrightarrow -13-9x=-x^2-37+2x \\ \Leftrightarrow x^2-11x+24=0 \\\text{We zoeken de oplossingen van } \color{blue}{x^2-11x+24=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (-11)^2-4.1.24 & &\\ & = 121-96 & & \\ & = 25 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-(-11)-\sqrt25}{2.1} & & = \frac{-(-11)+\sqrt25}{2.1} \\ & = \frac{6}{2} & & = \frac{16}{2} \\ & = 3 & & = 8 \\ \\ V &= \Big\{ 3 ; 8 \Big\} & &\end{align} \\ -----------------\)
12. \((5x+2)(4x+1)-x(18x+4)=4\\ \Leftrightarrow 20x^2+5x+8x+2 -18x^2-4x-4=0 \\ \Leftrightarrow 2x^2+3x-2=0 \\\text{We zoeken de oplossingen van } \color{blue}{2x^2+3x-2=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (3)^2-4.2.(-2) & &\\ & = 9+16 & & \\ & = 25 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-3-\sqrt25}{2.2} & & = \frac{-3+\sqrt25}{2.2} \\ & = \frac{-8}{4} & & = \frac{2}{4} \\ & = -2 & & = \frac{1}{2} \\ \\ V &= \Big\{ -2 ; \frac{1}{2} \Big\} & &\end{align} \\ -----------------\)