حل المعادلات التالية بالتحليل بالتجربة:

$y^2+16y+63=0$

$\begin{array}{rl}{y}^2+16y+63=0& \Rightarrow (y+9)(y+7)=0\\ & \Rightarrow y+9=0 \text{or }y+7=0\\ & \Rightarrow y=-9 \text{or }y=-7\Rightarrow s=\{-9,-7\}\end{array}$

تدرب وحل التمرينات

(1)- حل المعادلات التالية بالتحليل بالتجربة:

$x^2-15x+56=0$

$\begin{array}{rl}{x}^2-15x+56=0\Rightarrow (x-8)(x-7)=0& \Rightarrow \\ & \Rightarrow x-8=0 \text{or }x-7=0\\ & \Rightarrow x=8 \text{or }x=7\Rightarrow s=\{8,7\}\end{array}$

$y^2+16y+63=0$

$\begin{array}{rl}{y}^2+16y+63=0& \Rightarrow (y+9)(y+7)=0\\ & \Rightarrow y+9=0 \text{or }y+7=0\\ & \Rightarrow y=-9 \text{or }y=-7\Rightarrow s=\{-9,-7\}\end{array}$

$x^2+15x-16=0$

$\begin{array}{rl}& x^2+15x-16=0\Rightarrow (x-1)(x+16)=0\Rightarrow x-1=0 \text{or }x+16=0\\ & \Rightarrow x=1 \text{or }x=-16\Rightarrow s=\{1,-16\}\end{array}$

$y^2-y-42=0$

$\begin{array}{rl}{y}^2-y-42=0\Rightarrow (y-7)(y+6)=0& \Rightarrow y-7=0 \text{or }y+6=0\\ & \Rightarrow y=7 \text{or }y=-6\Rightarrow s=\{7,-6\}\end{array}$

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

$$\begin{gathered} x^2-4x+3=0\Rightarrow (x-3)(x-1)=0\Rightarrow x-3=0 \text{or }x-1=0 \\ \Rightarrow x=3 \text{or }x=1\Rightarrow s=\{3,1\} \end{gathered}$$

$y^2-6y-55=0$

$$\begin{gathered} y^2-6y-55-0\Rightarrow (y-11)(y+5)=0\Rightarrow y-11=0 \text{or }y+5=0 \\ \Rightarrow y=11 \text{or }y=-5\Rightarrow s=\{11,-5\} \end{gathered}$$

(2)- قطعة معدن مستطيلة الشكل ينقص عرضها بمقدار 2 m عن طولها ما بعدا القطعة المعدنية إذا كانت مساحتها 24m²؟

نفرض طول القطعة x لذا عرض القطعة x-2

$\begin{array}{rl}x(x-2)=24\Rightarrow x^2-2x-24-0\Rightarrow (x-6)(x+4)=0& \Rightarrow x-6=0 \text{or }x+4=0\\ & \Rightarrow x=6 \text{or }x=-4 \mathrm{يهمل}\end{array}$

$12x^2-20x+7=0$

$\begin{array}{rl}12x^2-20x+7=0\Rightarrow (2x-1)(6x-7)=0& \Rightarrow 2x-1=0 \text{or }6x-7=0\\ & \Rightarrow x=\frac{1}{2} \text{or }x=\frac{7}{6}\Rightarrow s=\{\frac{1}{2},\frac{7}{6}\}\end{array}$

$28+2z-8z^2=0$

$\begin{array}{rl}28+2z-8z^2=0\Rightarrow 14+z-4z^2& =0\\ \Rightarrow (2-z)(7+4z)=0& \Rightarrow 2-z=0 \text{or }7+4z=0\\ & \Rightarrow z=2 \text{or }z=\frac{-7}{4}\Rightarrow s=\{2,\frac{-7}{4}\}\end{array}$

$81-9x-12x^2=0$

$\begin{array}{rl}81+9x-12x^2=0& \Rightarrow 27-3x-4x^2=0\Rightarrow (3+x)(9-4x)=0\\ & \Rightarrow 3+x=0 \text{or }9-4x=0\Rightarrow x=-3 \text{or }x=\frac{9}{4}\Rightarrow s=\{-3,\frac{9}{4}\}\end{array}$

$50z^2+10z-4=0$

$\begin{array}{rl}50z^2-10z-4=0& \Rightarrow 25z^2-5z-2=0\Rightarrow (5z-1)(5z+2)=0\\ & \Rightarrow 5z-1=0 \text{or }5z+2=0\\ & \Rightarrow z=\frac{1}{5} \text{or }z=\frac{-2}{5}\Rightarrow s=\{\frac{1}{5},\frac{-2}{5}\}\end{array}$

(3)- صالة طعام ينقص طولها عن مثلي عرضها بمقدار 3 m ومساحتها 54 m²، ما أبعاد الصالة؟

نفرض عرض الصالة هو x لذا طول الصالة هو 2x-3

$\begin{array}{rl}x(2x-3)=54\Rightarrow 2x^2-3x=54& \Rightarrow 2x^2-3x-54=0\Rightarrow (2x+9)(x-6)=0\\ & \Rightarrow 2x+9=0 \text{or }x-6=0\\ & \Rightarrow x=\frac{-9}{5} \mathrm{يهمل} \text{or }x=6\end{array}$

عرض الصالة 6 m وطولها 9 m

(4)- جد مجموعة الحل للمعادلات التالية وتحقق من صحة الحل:

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

$$\begin{gathered} x^2-4x+3=0\Rightarrow (x-3)(x-1)=0\Rightarrow x-3=0 \text{or }x-1=0 \\ \Rightarrow x=3 \text{or }x=1\Rightarrow s=\{3,1\} \end{gathered}$$

التحقق:

$$\begin{gathered} \text{L.S}=3^2-4\left(3\right)+3=9-12+3=0=\text{R.S }\left(x=3\right) \\ \mathrm{L}.\mathrm{S}=1^2-4\left(1\right)+3=1-4+3=0=\text{R.S }\left(x=1\right) \end{gathered}$$

$y^2-9y-36=0$

$$\begin{gathered} y^2-9y-36=0\Rightarrow (y-12)(y+3)=0\Rightarrow y-12=0 \text{or }y+3=0 \\ \Rightarrow y=12 \text{or }y=-3\Rightarrow s=\{12,-3\} \end{gathered}$$

التحقق:

$\begin{array}{rl}& \text{L.S}=(12{)}^2-9(12)-36=144-108-36=0=\text{R.S }\left(y=12\right)\\ & \text{L.S}=(-3{)}^2-9(-3)-36=9+27-36=0=\text{R.S }\left(y=-3\right)\end{array}$

$4-26x+12x^2=0$

$\begin{array}{rl}4-26x+12x^2=0\Rightarrow 2-13x+6x^2=0& \Rightarrow (2-x)(1-6x)=0\\ & \Rightarrow 2\cdot x=0 \text{or }1-6x=0\\ & \Rightarrow x=2 \text{or }x=\frac{1}{6}\Rightarrow s=\{2,\frac{1}{6}\}\end{array}$

التحقق:

$$\begin{gathered} \mathrm{L}.S=4-26\left(2\right)+12\left(2^2\right)=4-52+48=0=\mathrm{R}.\mathrm{S} \left(\mathrm{x}=2\right) \\ \mathrm{L}.\mathrm{S}=4-26\left(\frac{1}{6}\right)+12{\left(\frac{1}{6}\right)}^2=4-\frac{13}{3}+\frac{1}{3}=\frac{12-13+1}{3}=0=\mathrm{R}.\mathrm{S} \left(\mathrm{x}=\frac{1}{6}\right) \end{gathered}$$

$80-38y+3y^2=0$

$\begin{array}{rl}80-38y+3y^2=0& \Rightarrow (10-y)(8-3y)=0\\ & \Rightarrow 10-y=0 \text{or }8-3y=0\\ & \Rightarrow y=10 \text{or }y=\frac{8}{3}\Rightarrow s=\{10,\frac{8}{3}\end{array}$

التحقق:

$\begin{array}{rl}& \text{L.S}=80-38\left(10\right)+3(10{)}^2=80-380+300=0=\text{R.S }\left(y=10\right)\\ & \text{L.S}=80-38\left(\frac{8}{3}\right)+3{\left(\frac{8}{3}\right)}^2=0=\text{R.S }\left(y=\frac{8}{3}\right)\end{array}$