Question

При каких а неравенство $x^2+ax-7a<0$ выполняется при всех х, таких что 1<x<2 ? 

Answer 1

 

$x^2 ax-7a<0, \ x^2 ax-7a=0, \ D=a^2 28a=a(a 28), \ Dgeq0, a(a 28)geq0, \ a(a 28)=0, \ a_1=-28, a_2=0, \ ain(-infty;-28]cup[0; infty), \ x_1=frac{-a-sqrt{a(a 28)}}{2}, x_2=frac{-a sqrt{a(a 28)}}{2}, \ frac{-a-sqrt{a(a 28)}}{2}<x$</var></p> <p> </p> <p><img src=[/tex]left { {{-a-sqrt{a^2+28a}leq2,} atop {-a+sqrt{a^2+28a}geq4;}} right. left { {{sqrt{a^2+28a}geq-(a+2),} atop {sqrt{a^2+28a}geq a+4;}} right. left { {{a^2+28ageq(a+2)^2,} atop {a^2+28ageq(a+4)^2;}} right. \ \ left { {{a^2+28ageq a^2+4a+4,} atop {a^2+28ageq a^2+8a+16;}} right. left { {{24ageq4,} atop {20ageq16;}} right. left { {{ageqfrac{1}{6},} atop {ageqfrac{4}{5};}} right. ageqfrac{4}{5}; \ ain[frac{4}{5};+infty)" title="x^2+ax-7a<0, \ x^2+ax-7a=0, \ D=a^2+28a=a(a+28), \ Dgeq0, a(a+28)geq0, \ a(a+28)=0, \ a_1=-28, a_2=0, \ ain(-infty;-28]cup[0;+infty), \ x_1=frac{-a-sqrt{a(a+28)}}{2}, x_2=frac{-a+sqrt{a(a+28)}}{2}, \ frac{-a-sqrt{a(a+28)}}{2}<x" title="left { {{-a-sqrt{a^2+28a}leq2,} atop {-a+sqrt{a^2+28a}geq4;}} right. left { {{sqrt{a^2+28a}geq-(a+2),} atop {sqrt{a^2+28a}geq a+4;}} right. left { {{a^2+28ageq(a+2)^2,} atop {a^2+28ageq(a+4)^2;}} right. \ \ left { {{a^2+28ageq a^2+4a+4,} atop {a^2+28ageq a^2+8a+16;}} right. left { {{24ageq4,} atop {20ageq16;}} right. left { {{ageqfrac{1}{6},} atop {ageqfrac{4}{5};}} right. ageqfrac{4}{5}; \ ain[frac{4}{5};+infty)" title="x^2+ax-7a<0, \ x^2+ax-7a=0, \ D=a^2+28a=a(a+28), \ Dgeq0, a(a+28)geq0, \ a(a+28)=0, \ a_1=-28, a_2=0, \ ain(-infty;-28]cup[0;+infty), \ x_1=frac{-a-sqrt{a(a+28)}}{2}, x_2=frac{-a+sqrt{a(a+28)}}{2}, \ frac{-a-sqrt{a(a+28)}}{2}<x" alt="left { {{-a-sqrt{a^2+28a}leq2,} atop {-a+sqrt{a^2+28a}geq4;}} right. left { {{sqrt{a^2+28a}geq-(a+2),} atop {sqrt{a^2+28a}geq a+4;}} right. left { {{a^2+28ageq(a+2)^2,} atop {a^2+28ageq(a+4)^2;}} right. \ \ left { {{a^2+28ageq a^2+4a+4,} atop {a^2+28ageq a^2+8a+16;}} right. left { {{24ageq4,} atop {20ageq16;}} right. left { {{ageqfrac{1}{6},} atop {ageqfrac{4}{5};}} right. ageqfrac{4}{5}; \ ain[frac{4}{5};+infty)" title="x^2+ax-7a<0, \ x^2+ax-7a=0, \ D=a^2+28a=a(a+28), \ Dgeq0, a(a+28)geq0, \ a(a+28)=0, \ a_1=-28, a_2=0, \ ain(-infty;-28]cup[0;+infty), \ x_1=frac{-a-sqrt{a(a+28)}}{2}, x_2=frac{-a+sqrt{a(a+28)}}{2}, \ frac{-a-sqrt{a(a+28)}}{2}<x" />

 

$x^2+ax-7a<0, \ x^2+ax-7a=0, \ D=a^2+28a=a(a+28), \ Dgeq0, a(a+28)geq0, \ a(a+28)=0, \ a_1=-28, a_2=0, \ ain(-infty;-28]cup[0;+infty), \ x_1=frac{-a-sqrt{a(a+28)}}{2}, x_2=frac{-a+sqrt{a(a+28)}}{2}, \ frac{-a-sqrt{a(a+28)}}{2}<x$

 

[tex]left { {{-a-sqrt{a^2+28a}leq2,} atop {-a+sqrt{a^2+28a}geq4;}} right. left { {{sqrt{a^2+28a}geq-(a+2),} atop {sqrt{a^2+28a}geq a+4;}} right. left { {{a^2+28ageq(a+2)^2,} atop {a^2+28ageq(a+4)^2;}} right. \ \ left { {{a^2+28ageq a^2+4a+4,} atop {a^2+28ageq a^2+8a+16;}} right. left { {{24ageq4,} atop {20ageq16;}} right. left { {{ageqfrac{1}{6},} atop {ageqfrac{4}{5};}} right. ageqfrac{4}{5}; \ ain[frac{4}{5};+infty)" />