Factorise x^2+9x+18 and hence solve x^2+9x+18=0

1 answer:

3 0

$x^2+9x+18\\\\9x=3x+6x\\18=3\cdot6\\\\x^2+9x+18=x^2+3x+6x+18=x(x+3)+6(x+3)=(x+3)(x+6)\\\\x^2+9x+18=0\iff(x+3)(x+6)=0\iff x+3=0\ or\ x+6=0\\\\x=-3\ or\ x=-6$

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Need help and explain please!!

lukranit [14]

Answer:

$x=-4\text{ and } x=3$

Step-by-step explanation:

We are given the second derivative:

$g''(x)=(x-3)^2(x+4)(x-6)$

And we want to find its inflection points.

To do so, we will first determine possible inflection points. These occur whenever g''(x) = 0 or is undefined.

Next, we will test values for the intervals. Inflection points occur if and only if the sign changes before and after the point.

So first, finding the zeros, we see that:

$0=(x-3)^2(x+4)(x-6)\Rightarrow x=-4, 3, 6$

So, we can draw the following number-line:

<----(-4)--------------(3)----(6)---->

Now, we will test values for the intervals x < -4, -4 < x < 3, 3 < x < 6, and x > 6.

Testing for x < -4, we can use -5. So:

$g^\prime^\prime(-5)=(-5-3)^2(-5+4)(-5-6)=704>0$

Since we acquired a positive result, g(x) is concave up for x < -4.

For -4 < x < 3, we can use 0. So:

$g^\prime^\prime(0)=(0-3)^2(0+4)(0-6)=-216$

Since we acquired a negative result, g(x) is concave down for -4 < x < 3.

And since the sign changed before and after the possible inflection point at x = -4, x = -4 is indeed an inflection point.

For 3 < x < 6, we can use 4. So:

$g^\prime^\prime(4)=(4-3)^2(4+4)(4-6)=-16$

Since we acquired a negative result, g(x) is concave down for 3 < x < 6.

Since the sign didn't change before and after the possible inflection point at x = 3 (it stayed negative both times), x = -3 is not a inflection point.

And finally, for x > 6, we can use 7. So:

$g^\prime^\prime(7)=(7-3)^2(7+4)(7-6)=176>0$