تدرب وحل التمرينات

(1)- حل المعادلات التالية بالمربع الكامل:

$x^2+24x+144=0$

$\begin{array}{rl}{x}^2+24x+144=0\Rightarrow (x{)}^2+24x+(12{)}^2=0& \Rightarrow (x+12{)}^2=0\\ & \Rightarrow x+12=0\Rightarrow x=-12\end{array}$

$y^2-20y+100=0$

$\begin{array}{rl}{y}^2-20y+100=0& \Rightarrow (x-10{)}^2=0\\ & \Rightarrow y-10=0\Rightarrow y=10\end{array}$

$y^2+4\sqrt{2}y+8=0$

$\begin{array}{rl}{y}^2+4\sqrt{2}y+8=0\Rightarrow & (y+2\sqrt{2}{)}^2=0\\ \Rightarrow y+2\sqrt{2}& =0\Rightarrow y=-2\sqrt{2}\end{array}$

$7-2\sqrt{7}z+z^2=0$

$\begin{array}{rl}7-2\sqrt{7}\mathrm{z}+{\mathrm{z}}^2=0\Rightarrow (\sqrt{7}-\mathrm{z}{)}^2& =0\\ & \Rightarrow \sqrt{7}-\mathrm{z}=0\Rightarrow \mathrm{z}=\sqrt{7}\end{array}$

$3y^2+36-12\sqrt{3}y=0$

$\begin{array}{rl}3y^2+36-12\sqrt{3}y=0& \Rightarrow 3y^2-12\sqrt{3}y+36=0\Rightarrow (\sqrt{3}y-6{)}^2=0\\ & \Rightarrow \sqrt{3}y-6=0\Rightarrow y=\frac{6}{\sqrt{3}}=2\sqrt{3}\end{array}$

$9z^2-10z+\frac{25}{9}=0$

$\begin{array}{rl}9z^2-10z+\frac{25}{9}=0& \Rightarrow {(3z-\frac{5}{3})}^2=0\\ & \Rightarrow 3z-\frac{5}{3}=0\Rightarrow z=\frac{5}{9}\end{array}$

(2)- حل المعادلات التالية بإكمال المربع:

$y^2+2\sqrt{3}y=3$

$\begin{array}{rl}& y^2-8y=24\Rightarrow y^2-8y+16=24+16\Rightarrow (y-4{)}^2=40\Rightarrow y-4=\pm 2\sqrt{10}\\ & \Rightarrow \{\begin{array}{c}y-4=2\sqrt{10}\Rightarrow y=4+2\sqrt{10}\\ or y-4=-2\sqrt{10}\Rightarrow y=4-2\sqrt{10}\Rightarrow s=\{4+2\sqrt{10},4-2\sqrt{10}\}\end{array}\end{array}$

$4z^2-12z-27=0$

$\begin{array}{rl}& 4z^2-12z-27=0\Rightarrow 4z^2-12z=27\Rightarrow z^2-3z=\frac{27}{4}\Rightarrow z^2-3z+\frac{9}{4}=\frac{27}{4}+\frac{9}{4}\\ & \Rightarrow {(z-\frac{3}{2})}^2=\frac{36}{4}\Rightarrow z-\frac{3}{2}=\pm \frac{6}{2}\\ & \Rightarrow \{\begin{array}{c}z-\frac{3}{2}=\frac{6}{2}\Rightarrow z=\frac{9}{2}\\ \text{or }z-\frac{3}{2}=\frac{-6}{2}\Rightarrow z=-\frac{-3}{2}\end{array}\\ & \Rightarrow s=\{\frac{9}{2},\frac{-3}{2}\}\end{array}$

$x^2-2x=0$

$\begin{array}{rl}{x}^2-2x=0\Rightarrow x^2-2x+1=1\Rightarrow (x-1{)}^2=1& \Rightarrow x-1=\pm 1\\ & \Rightarrow \{\begin{array}{c}x-1=1\Rightarrow x=2\\ \text{or }x-1=-1\Rightarrow x=0\Rightarrow s=\{2,0\}\end{array}\end{array}$

$y^2-8y=24$

$\begin{array}{rl}{y}^2-8y=24\Rightarrow y^2-8y+16& =24+16\Rightarrow (y-4{)}^2=40\Rightarrow y-4=\pm 2\sqrt{10}\\ \Rightarrow \{y-4& =2\sqrt{10}\Rightarrow y=4+2\sqrt{10}\\ or y-4& =-2\sqrt{10}\Rightarrow y=4-2\sqrt{10}\Rightarrow s=\{4+2\sqrt{10},4-2\sqrt{10}\}\end{array}$

$x^2-\frac{2}{3}x=4$

$\begin{array}{rl}{x}^2-\frac{2}{3}x=4\Rightarrow x^2-\frac{2}{3}x+\frac{1}{9}=4+& \frac{1}{9}\Rightarrow {(x-\frac{1}{3})}^2=\frac{37}{9}\\ \Rightarrow & x-\frac{1}{3}=\pm \frac{\sqrt{37}}{3}\\ \Rightarrow & \{x-\frac{1}{3}=\frac{\sqrt{37}}{3}\Rightarrow x=\frac{1+\sqrt{37}}{3}\\ & \text{or }x-\frac{1}{3}=\frac{-\sqrt{37}}{3}\Rightarrow x=\frac{1-\sqrt{37}}{3}\\ \Rightarrow & s=\{\frac{1+\sqrt{37}}{3},\frac{1-\sqrt{37}}{3}\}\end{array}$

$8y^2+16y-64=0$

$\begin{array}{rl}8y^2+16y-64=0\Rightarrow y^2+2y=8\Rightarrow y^2+2y+1=9& \Rightarrow (y+1{)}^2=9\Rightarrow y+1=\pm 3\\ \Rightarrow \{\begin{array}{l}y+1=3\Rightarrow y=2\\ or y+1=-3\Rightarrow y=-4\Rightarrow s=\{2,-4\}\end{array}& \end{array}$

(3)- حل المعادلات التالية بإكمال المربع، وجد الناتج بالتقريب لأقرب عدد صحيح:

$x^2-6x=15$

$\begin{array}{rl}{x}^2-6x=15\Rightarrow x^2-6x+9=15+9& \Rightarrow (x-3{)}^2=24\Rightarrow x-3\approx \pm 5\left(\sqrt{24}=\sqrt{25}=5\right)\\ & \Rightarrow \{\begin{array}{c}x-3\approx 5\Rightarrow x\approx 8\\ or x-3\approx -5\Rightarrow x\approx -2\end{array}\Rightarrow s=\{8,-2\}\end{array}$

$y(2y+28)=28$

$\begin{array}{rl}y(2y+28)=28& \Rightarrow 2y^2-28y=28\Rightarrow y^2-14y=14\\ \Rightarrow y^2-14y+49& =14+49\Rightarrow (y+7{)}^2=63\Rightarrow y+7\approx \pm 8 \left(\sqrt{63}\approx \sqrt{64}=8\right)\\ & \Rightarrow \{\begin{array}{c}y+7\approx 8\Rightarrow y\approx 1\\ or y+7\approx -8\Rightarrow y\approx -15\Rightarrow s=\{1,-15\}\end{array}\end{array}$

$z^2-10z+10=0$

$\begin{array}{rl}{\mathrm{z}}^2-10\mathrm{z}=-10& \Rightarrow {\mathrm{z}}^2-10\mathrm{z}+25=-10+25\\ & \Rightarrow (\mathrm{z}-5{)}^2=15\Rightarrow \mathrm{z}-5\approx \pm 4 \left(\sqrt{15}\approx \sqrt{16}=4\right)\\ & \Rightarrow \{\begin{array}{c}\mathrm{z}-5\approx 4\Rightarrow \mathrm{z}=9\\ \mathrm{or}-5\approx -4\Rightarrow \mathrm{z}\approx 1\Rightarrow \mathrm{s}=\{9,1\}\end{array}\end{array}$