Question

Please Help! I have no idea how to do this...
Which value of c makes the expression x2 + 7x + c a perfect square trinomial?

Answer 1

add (7/2)^2

Answer 2

@sourwing what do you mean. do you mean to plug it in to x?

Answer 3

no, c = (7/2)^2

Answer 4

@sourwing

a)49/4

b)-7/2

c)7/2

d)49

Answer 5

well (7/2)^2 = 49/4

Answer 6

a perfect square is of the form\[(x+a)(x+a) = x^2 + ax + ax + a^2\]\[=x^2+2ax+a^2\]

Comparing with your expression:
\[x^2+2ax+a^2\]\[x^2+7x+c\]
\[2ax = 7x\]\[2a=7\]\[a=\frac{7}{2}\]

and\[a^2=c\]\[(\frac{7}{2})^2 =c\]\[c = \frac{49}{4}\]

Answer 7

@whpalmer4 wow you're fast. Thank you! once again :D

Answer 8

Completing the square is a technique you'll often encounter when solving quadratic equations (such as the one from your other problem. A quick demo:

\[0 = 2x^2-6x-8\]We can factor out a 2 from everything without changing the solutions: we're just dividing the whole thing by 2
\[0 = x^2-3x-4\]Now we take the value of the coefficient of \(x\), which is \(-3\), divide it by \(2\) giving us \(-3/2\), and square it to get the final term in a perfect square: \(9/4\). We can't just add that to our equation without making some adjustment elsewhere, however! We have two basic options:

1) add it to both sides
2) add it to one side and also subtract it from the same side

Let's do #1

\[0+\frac{9}{4} = x^2-3x-4+\frac{9}{4}\]\[\frac{9}4+4 = x^2-3x+\frac{9}{4}\]Now the thing on the right is a perfect square, so we write it as one:\[\frac{9}{4} + 4 = (x-\frac{3}{2})^2\]Now we simplify the stuff on the left:
\[\frac{9}4+\frac{4}{4}*4 = (x-\frac{3}{2})^2\]\[\frac{25}{4} = (x-\frac{3}{2})^2\]Now we take the square root of both sides:
\[\pm\sqrt{\frac{25}{4}} = x-\frac{3}{2}\]\[\pm\frac{5}{2} = x-\frac{3}{2}\]Solve for \(x\) and we have\[x = \frac{3+5}{2} = 4\]\[x = \frac{3-5}{2} = -1\]

And as you recall, our original equation had those solutions.

Answer 9

@whpalmer4 Thank you! Do you use a the web to find these answers?

Answer 10

No, I use my brain :-)

Answer 11

With practice, you, too, can become similarly proficient.

Answer 12

@whpalmer4 Well you must be really really smart and thats a good thing :D thank you!

Answer 13

@whpalmer4 can you help me on another problem please

Answer 14

I've had a lot of practice. Much of the stuff in pre-college math isn't really difficult, it's just a matter of getting the right explanation, and working carefully.

Put up the problem, and we'll find out :-)

Answer 15

@whpalmer4 ok