1. Apr 6, 2005

DethRose

Hey

I have a test today and cant figure out how to solve certain quadratic equations.

We are doing solving the equations by completing the square using these steps:

1. divide each side by coefficient of x squared
2. rewrite the equation with the constant on right side
3. complete the square: add the square of one half of the coefficient of x to both sides
4. write the left side as a square and simplify the right side
5. equate the square root of the left side to the principle square root of the right side and its negative.
6. solve the 2 resulting equations.

Heres my problem:

2s squared+5s=3

heres what im doing

s squared+5/2s=2/3

but when you hald 5/2 and then square it you get a fraction...how are you supposed to factor a number with a fraction in it?

2. Apr 6, 2005

Data

You don't need to even think about the factorization. The thing added to $x$ in the factorization is always just half the coefficient of $x$ (including a negative sign, if the coefficient of $x$ has one!!!). You made a little mistake here anyways though, so:

$$2s^2 + 5s = 3 \Longrightarrow s^2 + \frac{5}{2}s = \frac{3}{2}$$

$$\Longrightarrow s^2 + \frac{5}{2}s + \frac{25}{16} = \frac{3}{2}+\frac{25}{16}$$

$$\Longrightarrow \left(s+\frac{5}{4}\right)^2 = \frac{49}{16}$$

you can finish...

3. Apr 6, 2005

DethRose

where did you get the 25/16 from?

4. Apr 6, 2005

DethRose

by the way the answer is x=-3,1/2

5. Apr 6, 2005

Data

$25/16$ is half of $5/2$ squared. If you solve from the point I left off, you'll get those answers~

6. Apr 6, 2005

DethRose

5/2 squared is 25/4 how can 25/16 be half of it???

im not trying to say you dont know what your doing lol i just am still totally lost lol

7. Apr 6, 2005

Data

Other way. You half it, then square it

8. Apr 6, 2005

DethRose

for final answer all i get is x=5/4+ square root of 49/16
x=5/4- square root of 49/16

9. Apr 6, 2005

DethRose

nevermind i got it :surprised :surprised

10. Apr 6, 2005

Data

Here's a little explanation as to why, since you seem confused:

When you square a binomial, say, $a+b$, you get the following:

$$(a+b)^2 = (a+b)(a+b) = a^2 + ab + ba + b^2 = a^2 + 2ab + b^2$$

now, what happens if $a = x$? Well, then this is

$$x^2 + 2bx + b^2.$$

since we put no restrictions on $b$, we find that a polynomial

$$cx^2 + dx + z^2$$

can be factored into a perfect square if and only if

$$z = \left(\frac{d}{2}\right) \Longrightarrow z^2 = \left(\frac{d}{2}\right)^2$$

and that in this case,

$$cx^2 + dx + z^2 = (x+z)^2.$$

so for your example, we want to do something to make the left side of

$$s^2 + \frac{5}{2}s = \frac{3}{2}$$

into a perfect square. Well, here, on the left side, using the letters from above, $c=1, \ d = 5/ 2, \ z = 0$. We need $z = (d/2) = 5/4 \Longrightarrow z^2 = (d/2)^2 = (5/4)^2$, so we add

$$\left( \frac{5/2}{2} \right)^2 = \left(\frac{5}{4}\right)^2 = \frac{25}{16}$$

to each side, and thus it factors to

$$(s+z)^2 = \left(s + \frac{5}{4}\right)^2.$$

Last edited: Apr 6, 2005
11. Apr 6, 2005

DethRose

thanks...ok i tried to do the same thing for another of the same type question but got the wrong answer again lol.

3y squared=3y+2

so i did

3y squared-3y=2
y squared-y=2/3
y squared-y-2/4=2/3-2/4
(y-2/4)squared=1/6
y=2/4 +or- square root of 1/6

but the answer is x= 1/6 (3 +or- square root of 33)

12. Apr 6, 2005

dextercioby

No.

$$(a-b)^{2}=a^{2}-2ab+b^{2}$$

Daniel.

13. Apr 6, 2005

DethRose

i tried a different way to do it still no luck

3y squared-3y=2
y squared-y=2/3
y squared-y+1/4=2/3+1/4
y+1/2=square root of 11/12

i still get the wrong answer :grumpy:

14. Apr 6, 2005

DethRose

anyone? lol

15. Apr 6, 2005

dextercioby

Of course

$$x^{2}-x+\frac{1}{4}=\left(x-\frac{1}{2}\right)^{2}$$

according to the formula i wrote above...

Daniel.

16. Apr 6, 2005

DethRose

ok i changed it to that but i still get the wrong answer

now i get x=1/2 +or- square root of 11/12

the answer is supposed to be 1/6(3+or- square root of 33)

17. Apr 6, 2005

dextercioby

I little algebra may show you that the two (apparently different) real numbers are one & the same...

Daniel.

18. Apr 6, 2005

Data

Your answer is correct. It's the same as theirs in a different form.

$$\frac{1}{2} \pm \sqrt{\frac{11}{12}} = \frac{1}{2} \pm \frac{\sqrt{11}}{2\sqrt{3}}$$

$$= \frac{1}{3}\left(\frac{3}{2} \pm \frac{3}{2}\frac{\sqrt{11}}{\sqrt{3}}\right) = \frac{1}{6}\left(3 \pm \frac{3\sqrt{11}}{\sqrt{3}}\right)$$

$$= \frac{1}{6}\left( 3 \pm \sqrt{3}\sqrt{11}\right) = \frac{1}{6}\left(3 \pm \sqrt{33}\right)$$

19. Apr 6, 2005

DethRose

i see it now...thanks data you were a huge help