Question

Solve the equation for x∈Z 

-x² +8x -14 ≥ 0

Answer 1

`-x^2 +8x -14 \geq 0\\ -(x^2-8x+14)\geq0\\ -(x^2-8x+16-2)\geq0\\ -((x-4)^2-2)\geq0\\ -(x-4)^2+2\geq0\\ -(x-4)^2\geq-2\\ (x-4)^2\leq2\\ x-4 \leq \sqrt2 \wedge x-4\geq-\sqrt2\\ x\leq 4+\sqrt2 \wedge x\geq4-\sqrt2\\ x\in[4-\sqrt2,4+\sqrt2]\\\\ x\in[4-\sqrt2,4+\sqrt2] \wedge x\in\mathbb{Z}\\ \boxed{x=\{3,4,5\}}`

Answer 2

`ax^2+bx+c=0\\\Delta=b^2-4ac\\if\ \Delta < 0\ then\ no\ solutions\\if\ \Delta=0\ then\ one\ solution:x=(-b)/(2a)\\if\ \Delta > 0\ then\ two\ solutions:x=(-b\pm\sqrt\Delta)/(2a)\\================================\\\\-x^2+8x-14\geq0\ \ \ \ |multiply\ both\ sides\ by\ (-1)\ \{change\ \geq\ on\ \leq\}\\\\x^2-8x+14\leq0\\\\a=1;\ b=-8;\ c=14\\\\\Delta=(-8)^2-4\cdot1\cdot14=64-56=8 > 0\\\sqrt\Delta=\sqrt8=√(4\cdot2)=\sqrt4\cdot\sqrt2=2\sqrt2`


`x_1=(-(-8)-2\sqrt2)/(2\cdot1)=(8-2\sqrt2)/(2)=(8)/(2)-(2\sqrt2)/(2)=\boxed{4-\sqrt2}\\\\x_2=(-(-8)+2\sqrt2)/(2\cdot1)=(8+2\sqrt2)/(2)=(8)/(2)+(2\sqrt2)/(2)=\boxed{4+\sqrt2}\\\\============================`


`ax^2+bx+c=0\\\\if\ a > 0\ then\ the\ parabola\ o pen\ up\\if\ a < 0\ then\ the\ parabola\ o pen\ down\\========================\\\\a=1 > 0-therefore\ o pen\ up\ (look\ at\ the\ picture)\\\\===============================\\\\Answer:x\in\left<4-\sqrt2;\ 4+\sqrt2\right>`



`Solutions\ in\ \mathbb{Z}:\\\\\sqrt2\approx1.4\\\\threfore:4-\sqrt2\approx4-1.4=2.6\ and\ 4+\sqrt2\approx4+1.4=5.4\\\\look\ at\ the\ second\ picture:x=3\ or\ x=4\ or\ x=5\ (x\in\{3;\ 4;\ 5\})`