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1 year ago

How can I solve 4x^2 + 12x > -x -3

Mathematics

1 answer:

Damm [24]1 year ago

5 0

He have the following:

$4x^2+12x>-x-3$

solving:

$\begin{gathered} 4x^2+12x+x+3>-x-3+x+3 \\ 4x^2+13x+3>0 \\ \text{factoring} \\ (4x+1)(x+3)>0 \\ (4x+1)>0\rightarrow4x+1-1>-1\rightarrow\frac{4x}{4}>-\frac{1}{4}\rightarrow x>-\frac{1}{4} \\ (x+3)$

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Someone please help.

Alexxandr [17]

Hi there!

$f(x)+f(x+1)=4f(x)$

There are 3 parts to this equation:

f(x)

f(x+1)

4f(x)

We must first determine these three parts separately.

1) f(x)

We're given that $f(x)=3^x$ :

⇒ $f(x)=3^x$ :

2) f(x+1)

Now, we must find f(x+1). To do so, add 1 to x in the original function $f(x)=3^x$ :

⇒ $f(x+1)=3^x^+^1$

3) 4f(x)

To find 4f(x), multiply the original function $f(x)=3^x$ by 4:

$4f(x)=4*3^x$ :

4) Put it all together

Now, plug each of the three parts into the equation $f(x)+f(x+1)=4f(x)$ :

$f(x)+f(x+1)=4f(x)$

$3^x+3^x^+^1=4*3^x\\3^x+3^x*3=4*3^x$

Factor the left side

$3^x*(1+3)=4*3^x$

Divide both sides by 3^x

$1+3=4\\4=4$

Because this equation is true, $f(x)+f(x+1)=4f(x)$ is therefore true.

I hope this helps!

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3 years ago