Mathematics

find the roots ? x² + 7x + 12 =0 x²-8x +12=0


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DestinyNicolle

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We have to remember one simple pattern [latex]\Delta =b^2-4ac[/latex] 1. [latex]x^2+7x+12=0[/latex] in this case [latex]b=7[/latex] [latex]a=1[/latex] [latex]c=12[/latex] so: [latex]\Delta =7^2-4*1*12[/latex] [latex]\Delta =49-48[/latex] [latex]\Delta =1[/latex] [latex]\Delta >0[/latex] it means that there are 2 roots [latex]\sqrt{\Delta} =1[/latex] [latex]x_{1}=\frac{-b+\sqrt{\Delta}}{2a}[/latex] [latex]x_{2}=\frac{-b-\sqrt{\Delta}}{2a}[/latex] [latex]x_{1}=\frac{-7+1}{2}[/latex] [latex]x_{1}=\frac{-6}{2}[/latex] [latex]x_{1}=-3[/latex] [latex]x_{2}=\frac{-7-1}{2}[/latex] [latex]x_{2}=\frac{-8}{2}[/latex] [latex]x_{2}=-4[/latex] Roots are [latex]x_1=-4[/latex] and [latex]x_2=-3[/latex] 2. [latex]x^2-8x +12=0[/latex] in this case [latex]a=1[/latex] [latex]b=-8[/latex] [latex]c=12[/latex] [latex]\Delta =(-8)^2-4*1*12[/latex] [latex]\Delta =64-48[/latex] [latex]\Delta =16[/latex] [latex]\Delta >0[/latex] it means that there are 2 roots [latex]\sqrt{\Delta}=4[/latex] [latex]x_{1}=\frac{-b+\sqrt{\Delta}}{2a}[/latex] [latex]x_{2}=\frac{-b-\sqrt{\Delta}}{2a}[/latex] [latex]x_{1}=\frac{8+4}{2}[/latex] [latex]x_{1}=\frac{12}{2}[/latex] [latex]x_{1}=6[/latex] [latex]x_{2}=\frac{8-4}{2}[/latex] [latex]x_{2}=\frac{4}{2}[/latex] [latex]x_{2}=2[/latex] Roots:[latex]x_1=6[/latex] and [latex]x_2=4[/latex]

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yassautumn

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[latex]x^2 + 7x + 12 =0\\ x^2+4x+3x+12=0\\ x(x+4)+3(x+4)=0\\ (x+3)(x+4)=0\\ x=-3 \vee x=-4\\\\ x^2-8x+12=0\\ x^2-6x-2x+12=0\\ x(x-6)-2(x-6)=0\\ (x-2)(x-6)=0\\ x=2 \vee x=6[/latex]

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