Gebruik de discriminant om volgende vierkantsvergelijkingen op te lossen
- \((3x-5)(-2x-5)-x(-7x+8)=28\)
- \(\frac{13}{12}x=-\frac{1}{2}x^2-\frac{1}{2}\)
- \(-6x=-\frac{1}{2}x^2-\frac{11}{2}\)
- \(\frac{1}{3}x^2-3x+\frac{8}{3}=0\)
- \((-4x+2)(-3x-4)-x(11x+4)=-24\)
- \(\frac{9}{5}x^2+\frac{7}{20}x-\frac{1}{5}=0\)
- \(10x^2-(9x+2)=2x(x-12)\)
- \(\frac{16}{45}x^2-\frac{8}{5}x+\frac{9}{5}=0\)
- \(-\frac{1}{3}x=-\frac{1}{12}x^2+\frac{8}{3}\)
- \(10x^2-(5x-16)=x(x+19)\)
- \(\frac{7}{12}x=-\frac{1}{2}x^2+2\)
- \(x(x+4)=2(x-2)\)
Gebruik de discriminant om volgende vierkantsvergelijkingen op te lossen
Verbetersleutel
- \((3x-5)(-2x-5)-x(-7x+8)=28\\
\Leftrightarrow -6x^2-15x+10x+25 +7x^2-8x-28=0 \\
\Leftrightarrow x^2+2x-3=0 \\\text{We zoeken de oplossingen van } \color{blue}{x^2+2x-3=0} \\ \\\begin{align}
D & = b^2 - 4.a.c & & \\
& = (2)^2-4.1.(-3) & &\\
& = 4+12 & & \\
& = 16 & & \\ \\
x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\
& = \frac{-2-\sqrt16}{2.1} & & = \frac{-2+\sqrt16}{2.1} \\
& = \frac{-6}{2} & & = \frac{2}{2} \\
& = -3 & & = 1 \\ \\ V &= \Big\{ -3 ; 1 \Big\} & &\end{align} \\ -----------------\)
- \(\frac{13}{12}x=-\frac{1}{2}x^2-\frac{1}{2} \\
\Leftrightarrow \frac{1}{2}x^2+\frac{13}{12}x+\frac{1}{2}=0 \\
\Leftrightarrow \color{red}{12.} \left(\frac{1}{2}x^2+\frac{13}{12}x+\frac{1}{2}\right)=0 \color{red}{.12} \\
\Leftrightarrow 6x^2+13x+6=0 \\\text{We zoeken de oplossingen van } \color{blue}{6x^2+13x+6=0} \\ \\\begin{align}
D & = b^2 - 4.a.c & & \\
& = (13)^2-4.6.6 & &\\
& = 169-144 & & \\
& = 25 & & \\ \\
x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\
& = \frac{-13-\sqrt25}{2.6} & & = \frac{-13+\sqrt25}{2.6} \\
& = \frac{-18}{12} & & = \frac{-8}{12} \\
& = \frac{-3}{2} & & = \frac{-2}{3} \\ \\ V &= \Big\{ \frac{-3}{2} ; \frac{-2}{3} \Big\} & &\end{align} \\ -----------------\)
- \(-6x=-\frac{1}{2}x^2-\frac{11}{2} \\
\Leftrightarrow \frac{1}{2}x^2-6x+\frac{11}{2}=0 \\
\Leftrightarrow \color{red}{2.} \left(\frac{1}{2}x^2-6x+\frac{11}{2}\right)=0 \color{red}{.2} \\
\Leftrightarrow x^2-12x+11=0 \\\text{We zoeken de oplossingen van } \color{blue}{x^2-12x+11=0} \\ \\\begin{align}
D & = b^2 - 4.a.c & & \\
& = (-12)^2-4.1.11 & &\\
& = 144-44 & & \\
& = 100 & & \\ \\
x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\
& = \frac{-(-12)-\sqrt100}{2.1} & & = \frac{-(-12)+\sqrt100}{2.1} \\
& = \frac{2}{2} & & = \frac{22}{2} \\
& = 1 & & = 11 \\ \\ V &= \Big\{ 1 ; 11 \Big\} & &\end{align} \\ -----------------\)
- \(\frac{1}{3}x^2-3x+\frac{8}{3}=0\\
\Leftrightarrow \color{red}{3.} \left(\frac{1}{3}x^2-3x+\frac{8}{3}\right)=0 \color{red}{.3} \\
\Leftrightarrow x^2-9x+8=0 \\\text{We zoeken de oplossingen van } \color{blue}{x^2-9x+8=0} \\ \\\begin{align}
D & = b^2 - 4.a.c & & \\
& = (-9)^2-4.1.8 & &\\
& = 81-32 & & \\
& = 49 & & \\ \\
x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\
& = \frac{-(-9)-\sqrt49}{2.1} & & = \frac{-(-9)+\sqrt49}{2.1} \\
& = \frac{2}{2} & & = \frac{16}{2} \\
& = 1 & & = 8 \\ \\ V &= \Big\{ 1 ; 8 \Big\} & &\end{align} \\ -----------------\)
- \((-4x+2)(-3x-4)-x(11x+4)=-24\\
\Leftrightarrow 12x^2+16x-6x-8 -11x^2-4x+24=0 \\
\Leftrightarrow x^2+4x+16=0 \\\text{We zoeken de oplossingen van } \color{blue}{x^2+4x+16=0} \\ \\\begin{align}
D & = b^2 - 4.a.c & & \\
& = (4)^2-4.1.16 & &\\
& = 16-64 & & \\
& = -48 & & \\ & < 0 \\V &= \varnothing \end{align} \\ -----------------\)
- \(\frac{9}{5}x^2+\frac{7}{20}x-\frac{1}{5}=0\\
\Leftrightarrow \color{red}{20.} \left(\frac{9}{5}x^2+\frac{7}{20}x-\frac{1}{5}\right)=0 \color{red}{.20} \\
\Leftrightarrow 36x^2+7x-4=0 \\\text{We zoeken de oplossingen van } \color{blue}{36x^2+7x-4=0} \\ \\\begin{align}
D & = b^2 - 4.a.c & & \\
& = (7)^2-4.36.(-4) & &\\
& = 49+576 & & \\
& = 625 & & \\ \\
x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\
& = \frac{-7-\sqrt625}{2.36} & & = \frac{-7+\sqrt625}{2.36} \\
& = \frac{-32}{72} & & = \frac{18}{72} \\
& = \frac{-4}{9} & & = \frac{1}{4} \\ \\ V &= \Big\{ \frac{-4}{9} ; \frac{1}{4} \Big\} & &\end{align} \\ -----------------\)
- \(10x^2-(9x+2)=2x(x-12) \\
\Leftrightarrow 10x^2-9x-2=2x^2-24x \\
\Leftrightarrow 8x^2+15x-2=0 \\\text{We zoeken de oplossingen van } \color{blue}{8x^2+15x-2=0} \\ \\\begin{align}
D & = b^2 - 4.a.c & & \\
& = (15)^2-4.8.(-2) & &\\
& = 225+64 & & \\
& = 289 & & \\ \\
x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\
& = \frac{-15-\sqrt289}{2.8} & & = \frac{-15+\sqrt289}{2.8} \\
& = \frac{-32}{16} & & = \frac{2}{16} \\
& = -2 & & = \frac{1}{8} \\ \\ V &= \Big\{ -2 ; \frac{1}{8} \Big\} & &\end{align} \\ -----------------\)
- \(\frac{16}{45}x^2-\frac{8}{5}x+\frac{9}{5}=0\\
\Leftrightarrow \color{red}{45.} \left(\frac{16}{45}x^2-\frac{8}{5}x+\frac{9}{5}\right)=0 \color{red}{.45} \\
\Leftrightarrow 16x^2-72x+81=0 \\\text{We zoeken de oplossingen van } \color{blue}{16x^2-72x+81=0} \\ \\\begin{align}
D & = b^2 - 4.a.c & & \\
& = (-72)^2-4.16.81 & &\\
& = 5184-5184 & & \\
& = 0 & & \\ x & = \frac{-b\pm \sqrt{D}}{2.a} & & \\
& = \frac{-(-72)}{2.16} & & \\
& = \frac{9}{4} & & \\V &= \Big\{ \frac{9}{4} \Big\} & &\end{align} \\ -----------------\)
- \(-\frac{1}{3}x=-\frac{1}{12}x^2+\frac{8}{3} \\
\Leftrightarrow \frac{1}{12}x^2-\frac{1}{3}x-\frac{8}{3}=0 \\
\Leftrightarrow \color{red}{12.} \left(\frac{1}{12}x^2-\frac{1}{3}x-\frac{8}{3}\right)=0 \color{red}{.12} \\
\Leftrightarrow x^2-4x-32=0 \\\text{We zoeken de oplossingen van } \color{blue}{x^2-4x-32=0} \\ \\\begin{align}
D & = b^2 - 4.a.c & & \\
& = (-4)^2-4.1.(-32) & &\\
& = 16+128 & & \\
& = 144 & & \\ \\
x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\
& = \frac{-(-4)-\sqrt144}{2.1} & & = \frac{-(-4)+\sqrt144}{2.1} \\
& = \frac{-8}{2} & & = \frac{16}{2} \\
& = -4 & & = 8 \\ \\ V &= \Big\{ -4 ; 8 \Big\} & &\end{align} \\ -----------------\)
- \(10x^2-(5x-16)=x(x+19) \\
\Leftrightarrow 10x^2-5x+16=x^2+19x \\
\Leftrightarrow 9x^2-24x+16=0 \\\text{We zoeken de oplossingen van } \color{blue}{9x^2-24x+16=0} \\ \\\begin{align}
D & = b^2 - 4.a.c & & \\
& = (-24)^2-4.9.16 & &\\
& = 576-576 & & \\
& = 0 & & \\ x & = \frac{-b\pm \sqrt{D}}{2.a} & & \\
& = \frac{-(-24)}{2.9} & & \\
& = \frac{4}{3} & & \\V &= \Big\{ \frac{4}{3} \Big\} & &\end{align} \\ -----------------\)
- \(\frac{7}{12}x=-\frac{1}{2}x^2+2 \\
\Leftrightarrow \frac{1}{2}x^2+\frac{7}{12}x-2=0 \\
\Leftrightarrow \color{red}{12.} \left(\frac{1}{2}x^2+\frac{7}{12}x-2\right)=0 \color{red}{.12} \\
\Leftrightarrow 6x^2+7x-24=0 \\\text{We zoeken de oplossingen van } \color{blue}{6x^2+7x-24=0} \\ \\\begin{align}
D & = b^2 - 4.a.c & & \\
& = (7)^2-4.6.(-24) & &\\
& = 49+576 & & \\
& = 625 & & \\ \\
x_1 & = \frac{-b-\sqrt{D}}{2.a} & x_2 & = \frac{-b+\sqrt{D}}{2.a} \\
& = \frac{-7-\sqrt625}{2.6} & & = \frac{-7+\sqrt625}{2.6} \\
& = \frac{-32}{12} & & = \frac{18}{12} \\
& = \frac{-8}{3} & & = \frac{3}{2} \\ \\ V &= \Big\{ \frac{-8}{3} ; \frac{3}{2} \Big\} & &\end{align} \\ -----------------\)
- \(x(x+4)=2(x-2) \\
\Leftrightarrow x^2+4x=2x-4 \\
\Leftrightarrow x^2+2x+4=0 \\\text{We zoeken de oplossingen van } \color{blue}{x^2+2x+4=0} \\ \\\begin{align}
D & = b^2 - 4.a.c & & \\
& = (2)^2-4.1.4 & &\\
& = 4-16 & & \\
& = -12 & & \\ & < 0 \\V &= \varnothing \end{align} \\ -----------------\)