# Hi, some help with year 11 algebra please

• Apr 17th 2013, 12:31 PM
dylandriscoll79
Hi, some help with year 11 algebra please
Ok so the question is as follows: solve for y

y^2 + 6y + 9 = 0

I don't know if I'm correct but so far this is what I've got
y^2 + 6y = -9
y^2 + y = -9
6
y^2 + y= -1.5

I don't know if there's a way to type to the second on here?? so ^2 will do for now
• Apr 17th 2013, 12:40 PM
dokrbb
Re: Hi, some help with year 11 algebra please
working with a second degree equation of the type $\displaystyle ay^2+ by + c = 0$, you normally use

$\displaystyle \y = \frac {(-b) \pm \sqrt{b^{2}-4ac}}{2a}$

the sign $\displaystyle \pm$ shows that you may have two(2) roots for this equation and this depends on the values of the discriminant $\displaystyle \sqrt{b^{2}-4ac}}$

now just plug in the values and do the math,

dokrbb
• Apr 18th 2013, 08:24 PM
ibdutt
Re: Hi, some help with year 11 algebra please
This particular question you can also do by splitting the middle term. the algorithm is
for ax^2+ bx + c = 0; find the product ac which in this case is 9. next find such factors of this product such that their sum is b, the coefficient of x. in this case 9 = 3 x 3 and 3 + 3 = 6
rewrite the equation with middle term as the sum.
i.e., y^2+ 3y + 3y = 9=0
Now take common and proceed to factorize.
• Apr 18th 2013, 08:34 PM
takatok
Re: Hi, some help with year 11 algebra please
dokrbb is correct that the quadratic equation above will always work, sometimes you can factor it for a much less messy approach. Which if this was a HW problem is the approach wanted here.

$\displaystyle y^{2}+6y+9 = (y+3)(y+3) = 0$

So for that to equal zero, either y+3 = 0 , or y+3 = 0 (Most of the time the factors will be different.. but in this case they are the same so we'll get only 1 answer:
$\displaystyle y=-3$