Is (x+0.5)^2 a Perfect Square?

[laura11]

y=7x^2-28x+3


so far I've got...
y-7(x^2-4x)+3
then i divided 4/2=2
and then did 2^2 = 4
sooo
y=7(x^2-4x+4-4)+3
and that's as far as i can get

 

[HallsofIvy]

You've done the hard work, now just use a(b+ c)= ab+ ac
$7(x^2- 4x+ 4- 4)+ 3= 7(x^2- 4x+ 4)+ 7(-4)+ 3$

 

[laura11]

im supposed to end up in the form y=a(x-h)^2+k though

 

[zooxanthellae]

The solution HallsofIvy gave does in fact end up in such a form. To find it, consider: what does $$x^2 - 4x + 4$$ equal? In other words, how can you get from $$x^2 - 4x + 4$$ to (x-h)^2? Similarly, you can get from $$7(-4) + 3$$ to k.

 

[laura11]

did kind of trial and error and i guess it equals (x-2)^2
but how do you actually figure that out?

 

[eumyang]

You have to use the square of a binomial pattern:
$$(a - b)^2 = a^2 - 2ab + b^2$$.

And ffter completing the square a number of times you'll start remembering the perfect square trinomials:
$$x^2 - 2x + 1 = (x - 1)^2$$
$$x^2 - 4x + 4 = (x - 2)^2$$
$$x^2 - 6x + 9 = (x - 3)^2$$
$$x^2 - 8x + 16 = (x - 4)^2$$
$$x^2 - 10x + 25 = (x - 5)^2$$
... and so on.


69

 

[laura11]

ok sooo I've got y=7x^2-28x+3
= 7(x^2-4x)+3
=7(x^2-4x+4-4)+3
=7[(x-2)^2-4)]+3
=7(x-2)^2-28+3
=7(x-2)^2-25

so i thought that was the whole answer...
but the answer in the book says the answer is 7[(x-2)^2-25/7]
what did i do wrong?

 

[HallsofIvy]

You did nothing wrong- but you seem to have forgotten your basic algebra:

7a+ b= 7(a+ b/7) (distributive law)

 

[laura11]

aahhh ok thanks to everyone!

 

[laura11]

shoot...
k I am on the next one and its

3x^2+3x+5
soo i have =3(x^2+x)+5
=3(x^2+x+.25-.25)+5

but what do i do when its not a perfect square

 

[Dick]

x^2+x+0.25 is a perfect square. It's (x+0.5)^2. Isn't it? You do exactly the same things you did before.