Kumonekta sa mga eksperto at makakuha ng mga sagot sa IDNStudy.com. Alamin ang mga detalyadong sagot sa iyong mga tanong mula sa aming malawak na kaalaman sa mga eksperto.
Sagot :
[tex]1.)\\\\x^2 - 2x + 35 = 0\\\\a=1, \ \ b=-2, \ \ c=35 \\\\x_{1}=\frac{-b-\sqrt{b^2-4ac}}{2a}= \frac{2-\sqrt{ (-2)^2-4 \cdot 1\cdot 35}}{2 }= \frac{2-\sqrt{4-140}}{2 }= \frac{2-\sqrt{-136}}{2 }=\\\\\frac{2-\sqrt{4*34}i}{2 }=\frac{2-2\sqrt{ 34}i}{2 }=\frac{2(1- \sqrt{ 34}i)}{2 }=1- \sqrt{ 34}i[/tex]
[tex]x_{2}=\frac{-b+\sqrt{b^2-4ac}}{2a}= \frac{2+\sqrt{ (-2)^2-4 \cdot 1\cdot 35}}{2 } =\frac{2-2\sqrt{ 34}i}{2 }=\frac{2(1+ \sqrt{ 34}i)}{2 }=1- \sqrt{ 34}i \\\\If\ \ b^{2}-4ac<0, \ then \ roots \ are \ imaginary \ (non-real)[/tex]
[tex]2.)\\\\x^2 + 2x =48\\\\x^2 + 2x -48=0\\\\a=1, \ \ b= 2, \ \ c=-48 \\\\x_{1}=\frac{-b-\sqrt{b^2-4ac}}{2a}= \frac{-2-\sqrt{ 2^2-4 \cdot 1\cdot (-48)}}{2 }= \frac{-2-\sqrt{4+ 96}}{2 }= \frac{-2-\sqrt{100}}{2 }= \\\\=\frac{-2-10}{2 }=\frac{-12 }{2 }=-6[/tex]
[tex]x_{2}=\frac{-b+\sqrt{b^2-4ac}}{2a}= \frac{-2+\sqrt{ 2^2-4 \cdot 1\cdot (-48)}}{2 }= =\frac{-2+10}{2 }=\frac{8 }{2 }=-4[/tex]
[tex]3.) \\\\4x + 32 = -x^2\\\\ x^2 +4x+32=0\\\\a=1, \ \ b= 4, \ \ c=32 \\\\x_{1}=\frac{-b-\sqrt{b^2-4ac}}{2a}= \frac{-4-\sqrt{ 4^2-4 \cdot 1\cdot 32}}{2 }= \frac{-4-\sqrt{16-128}}{2 }= \frac{-4-\sqrt{-112}}{2 }= \\\\=\frac{-4- \sqrt{16*7}i}{2 }= \frac{-4- 4\sqrt{ 7}i}{2 }= \frac{2(-2- 2\sqrt{ 7}i)}{2 }= -2- 2\sqrt{ 7}i[/tex]
[tex]x_{2}=\frac{-b+\sqrt{b^2-4ac}}{2a}= \frac{-4+\sqrt{ 4^2-4 \cdot 1\cdot 32}}{2 }= \frac{2(-2+ 2\sqrt{ 7}i)}{2 }= -2+ 2\sqrt{ 7}i \\\\If\ \ b^{2}-4ac<0, \ then \ roots \ are \ imaginary \ (non-real)[/tex]
[tex]x_{2}=\frac{-b+\sqrt{b^2-4ac}}{2a}= \frac{2+\sqrt{ (-2)^2-4 \cdot 1\cdot 35}}{2 } =\frac{2-2\sqrt{ 34}i}{2 }=\frac{2(1+ \sqrt{ 34}i)}{2 }=1- \sqrt{ 34}i \\\\If\ \ b^{2}-4ac<0, \ then \ roots \ are \ imaginary \ (non-real)[/tex]
[tex]2.)\\\\x^2 + 2x =48\\\\x^2 + 2x -48=0\\\\a=1, \ \ b= 2, \ \ c=-48 \\\\x_{1}=\frac{-b-\sqrt{b^2-4ac}}{2a}= \frac{-2-\sqrt{ 2^2-4 \cdot 1\cdot (-48)}}{2 }= \frac{-2-\sqrt{4+ 96}}{2 }= \frac{-2-\sqrt{100}}{2 }= \\\\=\frac{-2-10}{2 }=\frac{-12 }{2 }=-6[/tex]
[tex]x_{2}=\frac{-b+\sqrt{b^2-4ac}}{2a}= \frac{-2+\sqrt{ 2^2-4 \cdot 1\cdot (-48)}}{2 }= =\frac{-2+10}{2 }=\frac{8 }{2 }=-4[/tex]
[tex]3.) \\\\4x + 32 = -x^2\\\\ x^2 +4x+32=0\\\\a=1, \ \ b= 4, \ \ c=32 \\\\x_{1}=\frac{-b-\sqrt{b^2-4ac}}{2a}= \frac{-4-\sqrt{ 4^2-4 \cdot 1\cdot 32}}{2 }= \frac{-4-\sqrt{16-128}}{2 }= \frac{-4-\sqrt{-112}}{2 }= \\\\=\frac{-4- \sqrt{16*7}i}{2 }= \frac{-4- 4\sqrt{ 7}i}{2 }= \frac{2(-2- 2\sqrt{ 7}i)}{2 }= -2- 2\sqrt{ 7}i[/tex]
[tex]x_{2}=\frac{-b+\sqrt{b^2-4ac}}{2a}= \frac{-4+\sqrt{ 4^2-4 \cdot 1\cdot 32}}{2 }= \frac{2(-2+ 2\sqrt{ 7}i)}{2 }= -2+ 2\sqrt{ 7}i \\\\If\ \ b^{2}-4ac<0, \ then \ roots \ are \ imaginary \ (non-real)[/tex]
Salamat sa iyong pakikilahok. Patuloy na magbahagi ng iyong mga ideya at kasagutan. Ang iyong kaalaman ay mahalaga sa ating komunidad. Para sa mabilis at eksaktong mga solusyon, isipin ang IDNStudy.com. Salamat sa iyong pagbisita at sa muling pagkikita.