The question is this:

$y= \frac{1}{2}(x+3)^2 - 4 \\ x+3y = -27$

I tried solving it but this is what i got and i don't know what went wrong! for the second equation and i got this $y = -\frac{1}{2}x - 9$

and for the first one

$y = \frac{1}{2}(x^2 + 3x + 3x + 9) - 4 \\ y = \frac{1}{2}(x^2 + 6x + 9)-4 \\ y = \frac{1}{2} x^2 + 3x + \frac{9}{2} - 4 \\ y = \frac{1}{2} x^2 + 3x + \frac{1}{2} $

and i substituted those equations so i got $\frac{1}{2}x^2 + \frac{7}{2}x + \frac{17}{2}$ am i supposed to find the solutions of this now? I think something went wrong or i used the wrong formula to solve this?

How do i go on to solve this?? and on the question it said ti look at the discriminant of the new quadratic formed after substitution b4 finding the roots but i didn't use the quadratic formula?


Your error is to have $y = -\frac{1}{2}x - 9$

It should be $y = -\frac{1}{3}x - 9$

Your other equation came (correctly, I believe) to $y=\frac 12 x^2+3x+\frac 12$ so equate the two:

$\frac 12 x^2+3x+\frac 12= -\frac{1}{3}x - 9$

I would multiply by 6 to get:

$3 x^2+18x+3= -2x - 54$


$3 x^2+20x+57= 0$

As they say, check the discriminant to see if there are no real roots, exactly one real root or two distinct roots.

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  • $\begingroup$ Ohh thanks a lot for the answer...than how do i go on to solve it? $\endgroup$ – MATH ASKER Sep 20 '15 at 23:30

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