Can someone help me out here?

Mathematics

1 answer:

Tomtit [17]2 years ago

5 0

What exactly are you trying to find?

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Solve the equation for x∈Z -x² +8x -14 ≥ 0

kenny6666 [7]

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

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

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

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