Mathematics, 09.11.2020 17:00 granthazenp5e9mj

Determine whether each trinomial is a perfect square or not.. 1.A2+2a+1 2.4+4a+a2 3.A2-3a+9 4.4a2+24a+36 5.A4-10a+25 6.25a2+20a+9 7.A2b2-14ab+49 8.A2-2a-1 9.16a2+24a+36 10.10+6a+a2

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Answers

1) Perfect square, 2) Perfect square, 3) Not a perfect square, 4) Perfect square, 5) Perfect square, 6) Not a perfect square, 7) Perfect square, 8) Not a perfect square, 9) Not a perfect square, 10) Not a perfect square.

Step-by-step explanation:

From Algebra we must remember that a trinomial is a perfect square if and only if:

$(a+b)^{2} = a^{2}+2\cdot a\cdot b + b^{2}$ , $\forall \,a,b\in\mathbb{R}$ (Eq. 1)

Now we proceed to prove each trinomial:

1) $a^{2}+2\cdot a +1$ :

If a^2 =a^2 and $b^{2} = 1$ , then $a = \pm a$ and $b =\pm 1$ . Therefore, $2\cdot a\cdot b = 2\cdot (\pm a)\cdot (\pm 1) = \pm 2\cdot a$ . In a nutshell, we conclude that this polynomial is a perfect square.

2) $4+4\cdot a +a^{2}$

If $a^{2} = 4$ and $b^{2} = a^{2}$ , then $a = \pm 2$ and $b = \pm a$ . Therefore, $2\cdot a \cdot b = 2\cdot (\pm 2)\cdot (\pm a) = \pm 4\cdot a$ . In a nutshell, we conclude that this polynomial is a perfect square.

3) $a^{2}-3\cdot a +9$

If $a^{2} = a^{2}$ and $b^{2} = 9$ , then $a = \pm a$ and $b = \pm 3$ . Therefore, $2\cdot a \cdot b = 2\cdot (\pm a)\cdot (\pm 3) = \pm 6\cdot a$ . In a nutshell, we conclude that this polynomial is not a perfect square.

4) $4\cdot a^{2}+24\cdot a + 36$

If $a^{2} = 4\cdot a^{2}$ and $b^{2} = 36$ , then $a = \pm 2\cdot a$ and $b = \pm 6$ . Therefore, $2\cdot a \cdot b = 2\cdot (\pm 2\cdot a)\cdot (\pm 6) = 24\cdot a$ . In a nutshell, we conclude that this polynomial is a perfect square.

5) $a^{4}-10\cdot a +25$

If $a^{2} = a^{4}$ and $b^{2} = 25$ , then $a = \pm a^{2}$ and $b = \pm 5$ . Therefore, $2\cdot a \cdot b = 2\cdot (\pm a^{2})\cdot (\pm 5) = \pm 10\cdot a^{2}$ . In a nutshell, we conclude that this polynomial is a perfect square.

6) $25\cdot a^{2}+20\cdot a + 9$

If $a^{2} = 25\cdot a^{2}$ and b = 9 , then $a = \pm 5\cdot a$ and $b = \pm 3$ . Therefore, $2\cdot a \cdot b = 2\cdot (\pm 5\cdot a)\cdot (\pm 3) = \pm 30\cdot a$ . In a nutshell, we conclude that this polynomial is not a perfect square.

7) $a^{2}\cdot b^{2}-14\cdot a \cdot b + 49$

If $a^{2} = a^{2}\cdot b^{2}$ and $b^{2} = 49$ , then $a = \pm a\cdot b$ and $b = \pm 7$ . Therefore, $2\cdot a\cdot b = 2\cdot (\pm a\cdot b)\cdot (\pm 7) = \pm 14\cdot a \cdot b$ . In a nutshell, we conclude that this polynomial is a perfect square.

8) $a^{2}-2\cdot a -1$

If $a^{2} = a^{2}$ and $b^{2} = -1$ , then $a = \pm a$ and $b = \pm i$ . Therefore, $2\cdot a \cdot b = \pm 2\cdot a \cdot i$ . In a nutshell, we conclude that this polynomial is not a perfect square.

9) $16\cdot a^{2} + 24\cdot a +36$

If $a^{2} = 16\cdot a^{2}$ and $b^{2} = \pm 36$ , then $a = \pm 4\cdot a$ and $b = \pm 6$ . Therefore, $2\cdot a \cdot b = 2\cdot (\pm 4\cdot a)\cdot (\pm 6) = \pm 48\cdot a$ . In a nutshell, we conclude that this polynomial is not a perfect square.

10) $10+6\cdot a +a^{2}$

If $a^{2} = 10$ and $b^{2} = a^{2}$ , then $a = \pm\sqrt{10}$ and $b = \pm a$ . Therefore, $2\cdot a \cdot b = \pm 2\cdot \sqrt{10}\cdot a$ . In a nutshell, we conclude that this polynomial is not a perfect square.

Answer from: Quest

i believe

step-by-step explanation:

from my conclusion c

Answer from: Quest

Okay so n is going to represent our first number. since the second number of five times greater than n, the second number will be represented as 5n. for us to figure out the two numbers we need to have the value of n first. this is going to be done by adding the two numbers together, n and 5n. it can be simplified as 6n, which the problem states is equal to 72. divide 72 by 6n and you can solve for n. n+5n=726n=72n=(72/6)n=12 now that we know the value of n, we have to multiply 5 times 12 which equals 60. so our two numbers/integers are 12 and 60

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