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4 years ago

For which of the following compound inequalities is there no solution? Select one: a. 8m+5>=5 and -4m+6<=-10 b. 6m<=-36

and m+24>20 c. -2m< 6 and 3m>24 d. m+9<8 and -2m>=10

Mathematics

1 answer:

enyata [817]4 years ago

6 0

$a.\\8m+5\geq5\
\ \ |-5\\8m\geq0\ \ \ |:8\\m\geq0\to m\in\left$

$b.\\6m\leq-36\ \ \ |:6\\m\leq-6\to m\in\left(-\infty;\
-6\right>\\\\m+24 > 20\ \ \ |-24\\m > -4\to m\in(-4;\
\infty)\\\\solution:\ m\in\left(-\infty;\ -6\right>\ \cap\ (-4;\ \infty)\to
m\in\O\to NO\ SOLUTION$

$c.\\-2m < 6\ \ \ |:(-2)\\m > -3\to m\in(-3;\ \infty)\\\\3m > 24\
\ \ |:3\\m > 8\to m\in(8;\ \infty)\\\\solution:\ m\in(-3;\ \infty)\ \cap\
(8;\ \infty)\to m\in(8;\ \infty)$

$d.\\m+9 < 8\ \ \ |-9\\m < -1\to m\in(-\infty;\ -1)\\\\-2m \geq10\ \
\ |:(-2)\\m\leq-5\to m\in\left(-\infty;-5\right>\\\\solution:\
m\in(-\infty;\ -1)\ \cap\ \left(-\infty;-5\right>\to
m\in\left(-\infty;-5\right>$

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$\large \boxed{\sf \ \ x=1, \ \ x=2+5i, \ \ x=2-5i \ \ }$

Step-by-step explanation:

Hello,

I assume that we are working in $\mathbb{C}$ , otherwise there is only one zero which is 1. Please consider the following.

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$x^2-4x+29=0 \ \text{ complete the square, 29 = 25 + 4} \\ \\ x^2-2\cdot 2 \cdot x+2^2+25=0 \\ \\ (x-2)^2=-25=(5i)^2 \ \text{ take the root } \\ \\x-2=\pm 5i \ \text{ add 2 } \\ \\ x = 2+5i \ \text{ or } \ x = 2-5i$

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