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3 years ago

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

Mathematics

2 answers:

kenny6666 [7]3 years ago

6 0

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

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

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

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

tensa zangetsu [6.8K]3 years ago

5 0

$-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\}}


$

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$\theta_{CAB}=128.316$

$\theta_{ABC}=25.842$

$\theta_{BCA}=25.842$

Step-by-step explanation:

A = (-3,0) , B = (1,3) , and C = (1,-3)

We're going to use the distance formula to find the length of the sides:

$r= \sqrt{(x_1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2}$

$AB= \sqrt{(-3-1)^2+(0-3)^2}=5$

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$CA= \sqrt{(1-(-3))^2+(-3-0)^2}=5$

we can use the cosine law to find the angle:

it is to be noted that:

the angle CAB is opposite to the BC.

the angle ABC is opposite to the AC.

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to find the CAB, we'll use:

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$\cos{\theta_{CAB}}=\dfrac{9^2-(5^2+5^2)}{-2(5)(5)}$

$\theta_{CAB}=\arccos{-\dfrac{0.62}}$

$\theta_{CAB}=128.316$

Although we can use the same cosine law to find the other angles. but we can use sine law now too since we have one angle!

To find the angle ABC

$\dfrac{\sin{\theta_{ABC}}}{AC}=\dfrac{\sin{CAB}}{BC}$

$\sin{\theta_{ABC}}=AC\left(\dfrac{\sin{CAB}}{BC}\right)$

$\sin{\theta_{ABC}}=5\left(\dfrac{\sin{128.316}}{9}\right)$

$\theta_{ABC}=\arcsin{0.4359}\right)$

$\theta_{ABC}=25.842$

finally, we've seen that the triangle has two equal sides, AB = CA, this is an isosceles triangle. hence the angles ABC and BCA would also be the same.

$\theta_{BCA}=25.842$

this can also be checked using the fact the sum of all angles inside a triangle is 180

$\theta_{ABC}+\theta_{BCA}+\theta_{CAB}=180$

$25.842+128.316+25.842$

$180$

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