What is the product of the binomals below (2x+3) (3x+3)

Mathematics

1 answer:

Korvikt [17]2 years ago

5 0

Answer:

$3*(2x+3)*(x+1)$

Step-by-step explanation:

pull like terms from the problem to re-arrange it into a product

$\left(2x+3\right)\left(3x+3\right)\\3x + 3 = 3 \cdot (x + 1)\\3 \cdot (2x + 3) \cdot (x + 1)\\$

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What will be the value of

madreJ [45]

The expression as given doesn't make much sense. I think you're trying to describe an infinitely nested radical. We can express this recursively by

$\begin{cases}a_1=\sqrt{42}\\a_n=\sqrt{42+a_{n-1}}\end{cases}$

Then you want to know the value of

$\displaystyle\lim_{n\to\infty}a_n$

if it exists.

To show the limit exists and that $a_n$ converges to some limit, we can try showing that the sequence is bounded and monotonic.

Boundedness: It's true that $a_1=\sqrt{42}\le\sqrt{49}=7$ . Suppose $a_k\le 7$ . Then $a_{k+1}=\sqrt{42+a_k}\le\sqrt{42+7}=7$ . So by induction, $a_n$ is bounded above by 7 for all $n$ .

Monontonicity: We have $a_1=\sqrt{42}$ and $a_2=\sqrt{42+\sqrt{42}}$ . It should be quite clear that $a_2>a_1$ . Suppose $a_k>a_{k-1}$ . Then $a_{k+1}=\sqrt{42+a_k}>\sqrt{42+a_{k-1}}=a_k$ . So by induction, $a_n$ is monotonically increasing.