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2 years ago

5

.help work.............................math

1 answer:

e-lub [12.9K]2 years ago

3 0

Answer:

Same

Step-by-step explanation:

x=2 y=2

x=3 y=3

x=4 y=4

x=5 y=5

and then graph it

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KonstantinChe [14]

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For what values of $x$ is $$\frac{x^2 + x + 3}{2x^2 + x - 6} \ge 0?$$ note: be thorough and explain why all points in your answe

Rus_ich [418]

To solve the inequality $\frac{x^2+x+3}{2x^2+x-6}\ge \:0$

$\mathrm{Factor\:the\:left\:hand\:side\:}\frac{x^2+x+3}{2x^2+x-6}:$

The numerator $x^2+x+3$ is not factorizable.

so factor the denominator $2x^2+x-6$ :

$2x^2+x-6=\left(2x^2-3x\right)+\left(4x-6\right)=x\left(2x-3\right)+2\left(2x-3\right)\\ \mathrm{Factor\:out\:}x\mathrm{\:from\:}2x^2-3x\mathrm{:\quad }x\left(2x-3\right)\\ \mathrm{Factor\:out\:}2\mathrm{\:from\:}4x-6\mathrm{:\quad }2\left(2x-3\right)\\$

Now take $\mathrm{Factor\:out\:common\:term\:}\left(2x-3\right)$

Then we get factor of the denominator as $\left(2x-3\right)\left(x+2\right)$

Thus $\frac{x^2+x+3}{2x^2+x-6}=\frac{x^2+x+3}{\left(x+2\right)\left(2x-3\right)}$

Now $\mathrm{Compute\:the\:signs\:of\:the\:factors\:of\:}\frac{x^2+x+3}{\left(x+2\right)\left(2x-3\right)}\\$

Signs of $x^2+x+3>0$

$\mathrm{Choosing\:ranges\:that\:satisfy\:the\:required\:condition:}\:\ge \:\:0$

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