I'm trying to find the vertex form of a standard form quadratic equation through completing the square, and I'm using this page to try and get through it: http://www.purplemath.com/modules/sqrvertx.htm

in step 5, it says 'convert the right-hand side to squared form'. what is this? the site itself uses the term seven times but has no explanation of it, and I really can't follow how they combined what they had into the 'squared form'.

the equation I'm working with is $2(x^2+3x+2.25)=y+9.5$, but I'd like a general explanation that I could try to apply myself.

do note that I'm aware this question is probably really dumb, and I'm most likely missing something basic or it's just an alternate term for something I know.

  • $\begingroup$ $x^2+3x+2.25=x^2+3x+\frac{9}{2}=x^2+2\left(\frac{3}{2}\right)+\left(\frac{3}{2}\right)^2=\left(x+\frac{3}{2}\right)^2$ I just put it in squared form. $\endgroup$ – John Wayland Bales Nov 29 '16 at 21:56

May as well post this as an answer.


I just put it in squared form.

When you correctly perform the steps of the 'complete the square' procedure, you are guaranteed to be able to put it in squared form the same way it was done above.

After completing the square your quadratic expression should always be in the form

$$ x^2+2tx +t^2=(x+t)^2$$

In your example you can see that $2t=3$ so $t=\frac{3}{2}$. Then you check to make sure that $\left(\frac{3}{2}\right)^2=2.25$ and you know you are on the right path.

  • $\begingroup$ thanks for the explanation. I'm still not 'getting it', so to speak, but I think at this point it's more of a lack of sleep issue than a lack of clarification issue. I'll try to have another look tomorrow and get back if I need any additional help or have questions on the matter. thank you! $\endgroup$ – tupsu Nov 29 '16 at 23:28

first solve for y.

y=2x^2 +6x-4.5

Then factorise out the value on front of the x^2

y= 2[x^2 +3x-2.25]

Once the coefficient of x^2 is one, you're ready for the next step. Half the value of the coefficient of x and square (x+ half the coefficient of x) and minus (the coefficient of x) squared. This may seem complicated but observe the example.

y= 2[(x-1.5)^2 -2.25-(1.5)^2]

Now just simplify and expand out again.

y= 2[(x-1.5)^2 -2.25-(1.5)^2]

y= 2[(x-1.5)^2 -4.5]

y= 2(x-1.5)^2 -9

aaaaand done. Tell me if I've made any mistakes and or if you have any questions.

  • $\begingroup$ This may not be the most efficient way to approach this particular question but it is how the general method works. $\endgroup$ – nford64 Nov 29 '16 at 22:11
  • $\begingroup$ doesn't match the answer I've gotten, including when double checking with my GDC - the final number should be -9.5. $\endgroup$ – tupsu Nov 29 '16 at 22:49

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