Matematyka
slime96
2017-06-25 18:18:35
Proszę rozwiązać równanie? x^{2}+8x= -14
Odpowiedź
mloda338
2017-06-25 21:14:34

I sposób: [latex] x^{2} +8x=-14[/latex] [latex] x^{2} +8x+14=0[/latex] [latex]Delta=8^{2}-4*1*14=64-56=8[/latex] [latex] sqrt{Delta} = sqrt{8} =2 sqrt{2}[/latex] [latex] x_{1} = frac{-8-2 sqrt{2} }{2} = -4- sqrt{2} [/latex] [latex] x_{2} = frac{-8+2 sqrt{2} }{2} = -4+ sqrt{2} [/latex] II sposób: [latex] x^{2} +8x=-14[/latex] [latex] x^{2} +8x+14=0[/latex] [latex] x^{2} +8x+16-2=0[/latex] [latex](x+4)^{2}-2=0[/latex] [latex](x+4+ sqrt{2})(x+4-sqrt{2} )=0[/latex] [latex] x_{1} = -4- sqrt{2} [/latex] [latex] x_{2} = -4+ sqrt{2} [/latex]

Poola96
2017-06-25 21:15:49

[latex]x^2+8x=-14 \\ x^2+8x+14=0 \\ delta=64-56=8 \\ sqrt{delta}=sqrt{8}=2sqrt{2} \\ x_1=frac{-8-2sqrt{2}}{2}=-4-sqrt{2} \\ x_2=frac{-8+2sqrt{2}}{2}=-4+sqrt{2}[/latex]

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