حل المعادلات التالية بإكمال المربع:

$x^2-10x-24=0$

$\begin{array}{rl}{x}^2-10x-24=0\Rightarrow x^2-10x+25=24+25& \Rightarrow (x-5{)}^2=49\Rightarrow x-5=\pm 7\\ & \Rightarrow \{\begin{array}{c}x-5=7\Rightarrow x=12\\ \text{or }x-5=-7\Rightarrow x=-2\Rightarrow s=\{12,-2\}\end{array}\end{array}$

تأكد من فهمك

(1)- حل المعادلات التالية بالمربع الكامل:

$x^2+12x+36=0$

$x^2+12x+36=0\Rightarrow (x+6{)}^2=0\Rightarrow x+6=0\Rightarrow x=-6$

${\mathrm{y}}^2-10y+25=0$

$y^2-10y+25=0\Rightarrow (y-5{)}^2=0\Rightarrow y-5=0\Rightarrow y=5$

$4x^2-4x+1=0$

$4x^2-4x+1=0\Rightarrow (2x-1{)}^2=0\Rightarrow 2x-1=0\Rightarrow x=\frac{1}{2}$

${\mathrm{y}}^2+2\sqrt{7}\mathrm{y}+7=0$

$y^2+2\sqrt{7}y+7=0\Rightarrow (y+\sqrt{7}{)}^2=0\Rightarrow y+\sqrt{7}=0\Rightarrow y=-\sqrt{7}$

$x^2+16x=-64$

$x^2-16x+64=0\Rightarrow (x+8{)}^2=0\Rightarrow x+8=0\Rightarrow x=-8$

$\frac{1}{16}-\frac{1}{2}x+x^2=0$

$\frac{1}{16}-\frac{1}{2}x+x^2=0\Rightarrow {(\frac{1}{4}-x)}^2=0\Rightarrow \frac{1}{4}-x=0\Rightarrow x=\frac{1}{4}$

(2)- حل المعادلات التالية بإكمال المربع:

$x^2-10x-24=0$

$\begin{array}{rl}{x}^2-10x-24=0\Rightarrow x^2-10x+25=24+25& \Rightarrow (x-5{)}^2=49\Rightarrow x-5=\pm 7\\ & \Rightarrow \{\begin{array}{c}x-5=7\Rightarrow x=12\\ \text{or }x-5=-7\Rightarrow x=-2\Rightarrow s=\{12,-2\}\end{array}\end{array}$

$y^2-3=2y$

$\begin{array}{rl}{y}^2-3=2y\Rightarrow y^2-2y=3\Rightarrow y^2-2y+1=3& +1\Rightarrow (y-1{)}^2=4\Rightarrow y-1=\pm 2\\ & \Rightarrow \{\begin{array}{c}y-1=2\Rightarrow y=3\\ \text{or }y-1=-2\Rightarrow y=-1\Rightarrow s=\{3,-1\}\end{array}\end{array}$

$4x^2-3x-16=0$

$\begin{array}{rl}& 4x^2-3x-16=0\Rightarrow 4x^2-3x=16\Rightarrow x^2-\frac{3}{4}x=4\Rightarrow x^2-\frac{3}{4}x+\frac{9}{64}=4+\frac{9}{64}\\ & \Rightarrow {(x-\frac{3}{8})}^2=\frac{265}{64}\Rightarrow x-\frac{3}{8}=\pm \frac{\sqrt{265}}{8}\Rightarrow \{x-\frac{3}{8}=\frac{\sqrt{265}}{8}\Rightarrow x=\frac{3+\sqrt{265}}{8}\\ & \text{or }x-\frac{3}{8}=\frac{-\sqrt{265}}{8}\Rightarrow x=\frac{3-\sqrt{265}}{8}\\ & \Rightarrow \mathrm{s}=1\frac{3+\sqrt{265}}{8},\frac{3-\sqrt{265}}{8} \mathrm{الحل} \mathrm{مجموعة}\end{array}$

$3y^2+2y=1$

$\begin{array}{c}\Rightarrow {\mathrm{y}}^2+2\mathrm{y}=1\Rightarrow {\mathrm{y}}^2+\frac{2}{3}y=\frac{1}{3}\Rightarrow {\mathrm{y}}^2+\frac{2}{3}y+\frac{4}{36}=\frac{1}{3}+\frac{4}{36}\Rightarrow {(y+\frac{2}{6})}^z=\frac{16}{36}\\ \Rightarrow y+\frac{2}{6}=\pm \frac{4}{6}\\ \begin{array}{l}\Rightarrow \{\begin{array}{c}y+\frac{2}{6}=\frac{4}{6}\Rightarrow y=\frac{1}{3}\\ \text{or }y+\frac{2}{6}=-\frac{4}{6}\Rightarrow y=-1\end{array}\\ \Rightarrow s=\{\frac{1}{3},-1\}\end{array}\end{array}$

$x^2-\frac{6}{5}x=\frac{1}{5}$

$\begin{array}{rl}{x}^2-\frac{6}{5}x=\frac{1}{5}\Rightarrow x^2-\frac{6}{5}x+\frac{9}{25}=\frac{1}{5}+\frac{9}{25}& \Rightarrow {(x-\frac{3}{5})}^2=\frac{14}{25}\\ & \Rightarrow x-\frac{3}{5}=\pm \frac{\sqrt{14}}{5}\\ & \begin{array}{rl}& \Rightarrow \{\begin{array}{c}x+\frac{3}{5}=\frac{\sqrt{14}}{5}\Rightarrow x=\frac{3+\sqrt{14}}{5}\\ \text{or }x+\frac{3}{5}=\frac{-\sqrt{14}}{5}\Rightarrow x=\frac{3-\sqrt{14}}{5}\end{array}\\ & \Rightarrow s=\{\frac{3+\sqrt{14}}{5},\frac{3-\sqrt{14}}{5}\}\end{array}\end{array}$

$5y^2+15y-30=0$

$\begin{array}{rl}5y^2+15y-30=0\Rightarrow y^2+3y=6& \Rightarrow y^2+3y+\frac{9}{4}=6+\frac{9}{4}\Rightarrow {(y+\frac{3}{2})}^2=\frac{33}{4}\\ & \Rightarrow y+\frac{3}{2}=\pm \frac{\sqrt{33}}{2}\\ & \begin{array}{ll}& \Rightarrow \{\begin{array}{l}y+\frac{3}{2}=\frac{\sqrt{33}}{2}\Rightarrow y=\frac{\sqrt{33}-3}{2}\\ \text{or }y+\frac{3}{2}=\frac{-\sqrt{33}}{2}\Rightarrow y=\frac{-\sqrt{33}-3}{2}\end{array}\\ & \Rightarrow \mathrm{s}=\{\frac{\sqrt{33}-3}{2},\frac{-\sqrt{33}-3}{2}\}\end{array}\end{array}$