Cos70⁰ cos 40⁰+ sin70⁰ sin40⁰
я в школе

3)Один из корней уравнения равен 2x^2+bx-18=0 равен 2. Найдите другой корень и коэффициент b.

2x²+bx-18=0 ⇔x² +(b/2)x-9 =0 .
{x₁+x₂ = - b/2 ; x₁*x₂ =-9 .
* * * По теореме Виета  * * *  для приведенного квадратного уравнения x²+px+q =0  верно  {x₁+x₂ = - р; x₁*x₂=q * * *
{2*x₂ =-9 ; 2+x₂ = - b/2 .⇔{x₂ =-4,5 ; b =-2(2+x₂) .⇔{x₂ =-4,5 ; b =5.

ответ : x₂ =-4,5  ; b=5.

4)Один из корней уравнения равен 3x^2+14x+c=0 равен -4. Найдите другой корень и коэффициент c.

3x²+14x+c=0⇔x² +(14/3)x+c/3 =0 .
{x₁+x₂ =-14/3 ; x₁*x₂ =c/3 .
{-4+x₂ =-14/3 ; c=3*(-4)*x₂ .⇔{ x₂ =-14/3+4 ; c =3*(-4)*x₂ .⇔
{x₂ =-2/3 ; c =3*(-4)*(-2/3) =8.

ответ : x₂ =-2/3  ; c =8 .

5)Разность корней уравнения x^2+13x+q=0 равна 5. Найдите коэффициент q.

x²+13x+q =0;
{x₁-x₂ = 5 ;  x₁+x₂ = -13 ; x₁*x₂ =q.⇔{x₁ = -4 ; x₂ = -9 ;q = (-4)*(-9) =36. 

ответ :  q =36 .

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