Question

Помогите решить неравенство (2x-7)(2x+7)_>6x-51

Answer 1

(2x-7)(2x+7)≥6x-51
4x²-49-6x+51≥0
4x²-6x+2≥0
2x²-3x+1≥0
2x²-3x+1=0
D=9-8=1
x₁=3-1= 2/4 =1/2
       4
x₂ =3+1 =1
        4
      +                  -              +
----------- 1/2 ----------- 1------------
\\\\\\\\\\\                         \\\\\\\\\\\\
x∈(-∞; 1/2]U[1; +∞)

Answer 2

$(2x-7)(2x+7) \geq 6x-51 \\ 4x^2+14x-14x-49 \geq 6x-51 \\ 4x^2-49 \geq 6x-51 \\ 4x^2-6x-49+51 \geq 0 \\ 4x^2-6x+2 \geq 0 \\ |:2 \\ 2x^2-3x+1 \geq 0 \\ D=9-8=1 \\ x_1= \frac{3-1}{4}= \frac{1}{2} \\ x_2= \frac{3+1}{4} =1 \\ (x-1)(x- \frac{1}{2}) \geq 0 \\ \left \{ {{x \leq \frac{1}{2} } \atop {x \geq 1}} \right.$