Make a perfect square

**Juggalomike** · November 20th 2009, 06:08 AM

I need to make 8x^2+36x a perfect square, which number would i add to it to do so?(this is for parametric sequences in calc 2, but i just need to know the number i use to make that part a perfect square)

**lvleph** · November 20th 2009, 06:24 AM

The idea here is that you want to write an equation such that $a^2x^2 + bx + c^2 = (ax + c)^2$ , so you have to come up with a way to solve for c and $c^2$ . In this case you want to know $c^2$ .One thing we know about squaring an polynomial of the form $(ax + c)$ is that b from above can be given by $b = 2ac$ . Therefore in your case you can solve for c and square it to get the value you need, i.e. solve for c in the following and then square the result
$36 = 2\sqrt{8}c\Rightarrow36^2 = 4\cdot 8 \cdot c^2$

**upender** · November 22nd 2009, 08:33 AM

here for the given problem if we add 41 it will make a perfect square

8x2+36x+41

**e^(i*pi)** · November 22nd 2009, 08:37 AM

Originally Posted by upender

here for the given problem if we add 41 it will make a perfect square

8x2+36x+41

For that equation $b^2-4ac=-16$ so it will be ok in $\mathbb{C}$ but I think the OP wants a real solution

I'd use the discriminant to find c. For it to be a perfect square we know that $b^2-4ac=0$

$36^2 = 4 \times 8 \times c$

$c = \frac{36^2}{32} = 40.5$

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