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skelet666 [1.2K]
2 years ago
10

I don't know how to find the answer for (3 + 6b + 3b^2). I'm still confused even after my teacher explained it. Could someone br

eak it down for me?
Mathematics
1 answer:
yarga [219]2 years ago
8 0

Answer:

steps below

Step-by-step explanation:

3 + 6b + 3b² = 3 (b² + 2b + 1) = 3 (b + 1)²

root: -1         if 3 + 6b + 3b² = 0               b = -1

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<em> </em><em> </em><em> </em><em> </em>

<em> </em><em> </em><em> </em><em> </em><em>N:</em><em>B </em><em>if</em><em> </em><em>you're </em><em>using</em><em> </em><em>decimal</em><em> </em><em>calculations</em><em>.</em><em> </em><em>T</em><em>he</em><em> </em><em>total</em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>a</em><em>mount</em><em> </em><em>of</em><em> </em><em>seeds</em><em> </em><em>will </em><em>be</em><em> </em><em>34</em><em>.</em><em>7</em><em>5</em>

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What number must you add to complete the square?<br> x^2 + 2x = -1
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\large\begin{array}{l} \textsf{You must add 1 to complete the square.}\\\\ \textsf{There is a very common special product: (square of a sum)}\\\\ \mathsf{(a+b)^2=a^2+2ab+b^2}\\\\\\ \textsf{Take the given equation:}\\\\ \mathsf{x^2+2x=-1\qquad(i)}\\\\\\ \textsf{The first term is a square (x squared)}\\\\ \textsf{The second term is a product of 2, x and 1.} \end{array}


\large\begin{array}{l}\\\\\\ \textsf{We need to find a proper value for b, so that if we add it to}\\\textsf{the left side of the equation, we get a perfect square.}\\\\ \textsf{So,}\\\\ \begin{array}{cccccccccc} \mathsf{a^2}&\!\!\!+\!\!\!&\mathsf{2}&\!\!\!\cdot\!\!\!&\mathsf{a}&\!\!\!\cdot\!\!\!&\mathsf{b}&\!\!\!+\!\!\!&\mathsf{b^2}&=\mathsf{(a+b)^2}\\\\ \mathsf{\downarrow}&&&\!\!\!\!\!&\downarrow&\!\!\!\!\!&\downarrow&&\downarrow\\\\ \mathsf{x^2}&\!\!\!+\!\!\!&\mathsf{2}&\!\!\!\cdot\!\!\!&\mathsf{x}&\!\!\!\cdot\!\!\!&\mathsf{1}&\!\!\!+\!\!\!&\mathsf{1^2}&=\mathsf{(x+1)^2} \end{array}\\\\\\ \textsf{Just by comparing the expressions above, we can conclude that}\\\textsf{b must be equal to 1, so we get a perfect square:}\\\\ \mathsf{(x+1)^2=x^2+2x+1} \end{array}


\large\begin{array}{l}\\\\\\ \textsf{Back to (i), adding 1 to both sides:}\\\\ \mathsf{x^2+2x+1=-1+1}\\\\ \boxed{\begin{array}{c} \mathsf{(x+1)^2=0} \end{array}} \end{array}


\large\begin{array}{l}\\\\\\ \textsf{If you want to solve this equation for x, you will find}\\\\ \mathsf{x+1=0}\\\\ \mathsf{x=-1}\quad\longleftarrow\quad\textsf{(solution)} \end{array}


If you're having problems understanding the answer, try seeing it through your browser: brainly.com/question/2112248


\large\begin{array}{l} \textsf{Any doubt? Please, comment below.}\\\\\\ \textsf{Best regards! :-)} \end{array}

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