$s= -5t^2+ 40t$.
Express $s$ in the form of $a(t-b)^2 + c$, where $a$, $b$ and $c$ are the constants.
$s = -5t(t-8)$. I have factorized it.
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1It will benefit you if you learn to format on this site. Formatting tips here. We also ask that you include your thoughts and efforts in every post. – Em. Mar 09 '16 at 23:33
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This question is about completing the square rather than the discriminant. Please read this tutorial on how to typeset mathematics on this site. Also, please edit your question to indicate what you have tried and where you are stuck so that you receive responses that address the specific difficulties you are encountering. – N. F. Taussig Mar 09 '16 at 23:33
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1s= -5t(t-8) I have factorized it... – Jeannine Mar 09 '16 at 23:34
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The first step in completing the square is to extract the factor of $-5$ to obtain $s = -5(t^2 - 8t)$. – N. F. Taussig Mar 09 '16 at 23:36
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How do you do that? – Jeannine Mar 09 '16 at 23:37
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Why do you add 80 outside the parentheses – Jeannine Mar 09 '16 at 23:58
2 Answers
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To transform the equation from standard form to vertex form, we must complete the square.
First, we write $$s = -5t^2 + 40t = -5(t^2 - 8t)$$ We need to transform $t^2 - 8t$ into a perfect square. Remember that $$(a + b)^2 = a^2 + 2ab + b^2$$ If we let $a = t$, then $2ab = 2bt = -8t \implies b = -4$. Thus, to create a perfect square inside the parentheses, we have to add $b^2 = (-4)^2 = 16$. However, $-5(t^2 - 8t + 16) \neq s$ since adding $16$ inside the parentheses adds $(-5)(16) = -80$ to $s$. To compensate, we must add $80$ outside the parentheses. Hence, \begin{align*} s & = -5t^2 + 40t\\ & = -5(t^2 - 8t)\\ & = -5(t^2 - 8t + 16) + 80\\ & = -5(t - 4)^2 + 80 \end{align*} where in the final step we have factored the expression $t^2 - 8t + 16$.
N. F. Taussig
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As I said in my answer, adding $16$ insides the parentheses adds $(-5)(16) = -80$ to $s$. In order to balance the equation, we must add $80$ outside the parentheses. As you can check, $-5(t^2 - 8t + 16) + 80 = -5(t^2 - 8t) = -5t^2 + 40t$. – N. F. Taussig Mar 10 '16 at 00:11
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First, take out a factor of $-5$ for $$s=-5(t^2-8t)$$ and then complete the square to get $$s=-5((t-4)^2-16).$$ In completing the square I divide the insides of the bracket by $t$, then half the number and deduct the square of it from the whole thing. It's not a very mathematically rigorous explanation but for a layman and for quickness I hope you can see what I have done. If you want to be sure you can work out $(t-4)^2-16$ and and it will be $t^2-8t.$ The rest should be pretty straight forward, just multiply through by that $-5$ to get $$a=-5\qquad b=4\qquad c=80$$
Nemon27
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A better guide on completing the square: https://www.mathsisfun.com/algebra/completing-square.html . Is that what you were questioning? – Nemon27 Mar 10 '16 at 00:04
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While you completed the square correctly, your explanation is incorrect. What you did was to divide $-8$ by $2$ to obtain $-4$, square $-4$ to obtain $16$, then adding $16$ inside the inner parentheses and adding $-16$ inside the outer parentheses in order to balance the equation. You certainly did not divide by $t$. – N. F. Taussig Mar 10 '16 at 00:19
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In my explanation I mean that I divide $t^2-8t$ by $t$ to get $t-8$. I then half the number (i.e. half $-8$ for $-4$) and deduct the square of it from the square of the $t-4$ meaning $(t-4)^2-16$. – Nemon27 Mar 10 '16 at 14:48