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Gebruik de discriminant om volgende vierkantsvergelijkingen op te lossen

1. \(10x^2-(12x-121)=x(x-78)\)
2. \(7x^2-(2x-21)=6x(x-2)\)
3. \(\frac{7}{9}x=-\frac{1}{3}x^2+\frac{16}{3}\)
4. \(6x^2-(5x-9)=2x(x-9)\)
5. \((5x-5)(-2x-5)-x(-19x+2)=24\)
6. \((-x+4)(4x+2)-x(-5x+12)=-1\)
7. \((-2x-2)(-2x-3)-x(-12x-12)=-115\)
8. \(-\frac{1}{2}x=-x^2-\frac{25}{4}\)
9. \((-5x-3)(x-1)-x(-21x+40)=-33\)
10. \(\frac{1}{48}x^2-\frac{1}{2}x+3=0\)
11. \(x(6x+31)=24(x+1)\)
12. \(\frac{9}{2}x=-x^2-\frac{121}{4}\)

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Gebruik de discriminant om volgende vierkantsvergelijkingen op te lossen

Verbetersleutel

1. \(10x^2-(12x-121)=x(x-78) \\ \Leftrightarrow 10x^2-12x+121=x^2-78x \\ \Leftrightarrow 9x^2+66x+121=0 \\\text{We zoeken de oplossingen van } \color{blue}{9x^2+66x+121=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (66)^2-4.9.121 & &\\ & = 4356-4356 & & \\ & = 0 & & \\ x & = \frac{-b\pm \sqrt{D}}{2.a} & & \\ & = \frac{-66}{2.9} & & \\ & = -\frac{11}{3} & & \\V &= \Big\{ -\frac{11}{3} \Big\} & &\end{align} \\ -----------------\)
2. \(7x^2-(2x-21)=6x(x-2) \\ \Leftrightarrow 7x^2-2x+21=6x^2-12x \\ \Leftrightarrow x^2+10x+21=0 \\\text{We zoeken de oplossingen van } \color{blue}{x^2+10x+21=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (10)^2-4.1.21 & &\\ & = 100-84 & & \\ & = 16 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-10-\sqrt16}{2.1} & & = \frac{-10+\sqrt16}{2.1} \\ & = \frac{-14}{2} & & = \frac{-6}{2} \\ & = -7 & & = -3 \\ \\ V &= \Big\{ -7 ; -3 \Big\} & &\end{align} \\ -----------------\)
3. \(\frac{7}{9}x=-\frac{1}{3}x^2+\frac{16}{3} \\ \Leftrightarrow \frac{1}{3}x^2+\frac{7}{9}x-\frac{16}{3}=0 \\ \Leftrightarrow \color{red}{9.} \left(\frac{1}{3}x^2+\frac{7}{9}x-\frac{16}{3}\right)=0 \color{red}{.9} \\ \Leftrightarrow 3x^2+7x-48=0 \\\text{We zoeken de oplossingen van } \color{blue}{3x^2+7x-48=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (7)^2-4.3.(-48) & &\\ & = 49+576 & & \\ & = 625 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-7-\sqrt625}{2.3} & & = \frac{-7+\sqrt625}{2.3} \\ & = \frac{-32}{6} & & = \frac{18}{6} \\ & = \frac{-16}{3} & & = 3 \\ \\ V &= \Big\{ \frac{-16}{3} ; 3 \Big\} & &\end{align} \\ -----------------\)
4. \(6x^2-(5x-9)=2x(x-9) \\ \Leftrightarrow 6x^2-5x+9=2x^2-18x \\ \Leftrightarrow 4x^2+13x+9=0 \\\text{We zoeken de oplossingen van } \color{blue}{4x^2+13x+9=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (13)^2-4.4.9 & &\\ & = 169-144 & & \\ & = 25 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-13-\sqrt25}{2.4} & & = \frac{-13+\sqrt25}{2.4} \\ & = \frac{-18}{8} & & = \frac{-8}{8} \\ & = \frac{-9}{4} & & = -1 \\ \\ V &= \Big\{ \frac{-9}{4} ; -1 \Big\} & &\end{align} \\ -----------------\)
5. \((5x-5)(-2x-5)-x(-19x+2)=24\\ \Leftrightarrow -10x^2-25x+10x+25 +19x^2-2x-24=0 \\ \Leftrightarrow 9x^2-2x+1=0 \\\text{We zoeken de oplossingen van } \color{blue}{9x^2-2x+1=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (-2)^2-4.9.1 & &\\ & = 4-36 & & \\ & = -32 & & \\ & < 0 \\V &= \varnothing \end{align} \\ -----------------\)
6. \((-x+4)(4x+2)-x(-5x+12)=-1\\ \Leftrightarrow -4x^2-2x+16x+8 +5x^2-12x+1=0 \\ \Leftrightarrow x^2-6x+9=0 \\\text{We zoeken de oplossingen van } \color{blue}{x^2-6x+9=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (-6)^2-4.1.9 & &\\ & = 36-36 & & \\ & = 0 & & \\ x & = \frac{-b\pm \sqrt{D}}{2.a} & & \\ & = \frac{-(-6)}{2.1} & & \\ & = 3 & & \\V &= \Big\{ 3 \Big\} & &\end{align} \\ -----------------\)
7. \((-2x-2)(-2x-3)-x(-12x-12)=-115\\ \Leftrightarrow 4x^2+6x+4x+6 +12x^2+12x+115=0 \\ \Leftrightarrow 16x^2+24x+121=0 \\\text{We zoeken de oplossingen van } \color{blue}{16x^2+24x+121=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (24)^2-4.16.121 & &\\ & = 576-7744 & & \\ & = -7168 & & \\ & < 0 \\V &= \varnothing \end{align} \\ -----------------\)
8. \(-\frac{1}{2}x=-x^2-\frac{25}{4} \\ \Leftrightarrow x^2-\frac{1}{2}x+\frac{25}{4}=0 \\ \Leftrightarrow \color{red}{4.} \left(x^2-\frac{1}{2}x+\frac{25}{4}\right)=0 \color{red}{.4} \\ \Leftrightarrow 4x^2-2x+25=0 \\\text{We zoeken de oplossingen van } \color{blue}{4x^2-2x+25=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (-2)^2-4.4.25 & &\\ & = 4-400 & & \\ & = -396 & & \\ & < 0 \\V &= \varnothing \end{align} \\ -----------------\)
9. \((-5x-3)(x-1)-x(-21x+40)=-33\\ \Leftrightarrow -5x^2+5x-3x+3 +21x^2-40x+33=0 \\ \Leftrightarrow 16x^2-32x+36=0 \\\text{We zoeken de oplossingen van } \color{blue}{16x^2-32x+36=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (-32)^2-4.16.36 & &\\ & = 1024-2304 & & \\ & = -1280 & & \\ & < 0 \\V &= \varnothing \end{align} \\ -----------------\)
10. \(\frac{1}{48}x^2-\frac{1}{2}x+3=0\\ \Leftrightarrow \color{red}{48.} \left(\frac{1}{48}x^2-\frac{1}{2}x+3\right)=0 \color{red}{.48} \\ \Leftrightarrow x^2-24x+144=0 \\\text{We zoeken de oplossingen van } \color{blue}{x^2-24x+144=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (-24)^2-4.1.144 & &\\ & = 576-576 & & \\ & = 0 & & \\ x & = \frac{-b\pm \sqrt{D}}{2.a} & & \\ & = \frac{-(-24)}{2.1} & & \\ & = 12 & & \\V &= \Big\{ 12 \Big\} & &\end{align} \\ -----------------\)
11. \(x(6x+31)=24(x+1) \\ \Leftrightarrow 6x^2+31x=24x+24 \\ \Leftrightarrow 6x^2+7x-24=0 \\\text{We zoeken de oplossingen van } \color{blue}{6x^2+7x-24=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (7)^2-4.6.(-24) & &\\ & = 49+576 & & \\ & = 625 & & \\ \\ x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\ & = \frac{-7-\sqrt625}{2.6} & & = \frac{-7+\sqrt625}{2.6} \\ & = \frac{-32}{12} & & = \frac{18}{12} \\ & = \frac{-8}{3} & & = \frac{3}{2} \\ \\ V &= \Big\{ \frac{-8}{3} ; \frac{3}{2} \Big\} & &\end{align} \\ -----------------\)
12. \(\frac{9}{2}x=-x^2-\frac{121}{4} \\ \Leftrightarrow x^2+\frac{9}{2}x+\frac{121}{4}=0 \\ \Leftrightarrow \color{red}{4.} \left(x^2+\frac{9}{2}x+\frac{121}{4}\right)=0 \color{red}{.4} \\ \Leftrightarrow 4x^2+18x+121=0 \\\text{We zoeken de oplossingen van } \color{blue}{4x^2+18x+121=0} \\ \\\begin{align} D & = b^2 - 4.a.c & & \\ & = (18)^2-4.4.121 & &\\ & = 324-1936 & & \\ & = -1612 & & \\ & < 0 \\V &= \varnothing \end{align} \\ -----------------\)