0

Is here anybody who can help me with mathematical adjustment between equation 8 and 9 in the picture?

Ignore the text between equations

Badam Baplan
  • 8,688
Miroslav Holub
  • 15
  • Welcome to maths SE. Could you edit your post to include the equation instead of a picture? MathJax will help you typeset your equation. Also, could you include your thoughts? – Alain Remillard Dec 13 '19 at 18:24

2 Answers2

0

It looks like they used the quadratic formula...

nomen
  • 2,707
  • 1
    This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post. - From Review – Badam Baplan Dec 13 '19 at 18:43
  • ^^ I think, as is, this is more appropriate as a comment. To qualify as an answer I think you should elaborate a little bit.... show a couple steps? – Badam Baplan Dec 13 '19 at 18:45
  • Did the OP show steps? – nomen Dec 13 '19 at 18:52
  • The OP was essentially asking how to get the canonical parabolic form from a quadratic equation. There's an intermediate step that the text in the image didn't spell out. Seemed to me like the OP just needed it to be made explicit. – Badam Baplan Dec 13 '19 at 19:35
0

This is an instance of completing the square.

Note that $$(x-\frac{CF_1}{2CF_2})^2 = x^2 - \frac{CF_1}{CF_2}x + \frac{CF_1^2}{4CF_2^2}$$

Recognize that the right hand side of this equation is just the right hand side of $(8)$ + $\frac{CF_1^2}{4CF_2^2}$.

Thus if you add $\frac{CF_1^2}{4CF_2^2}$ to both sides of the equation in $(8)$, you get that $$\frac{y - CF_0}{CF_2} + \frac{CF_1^2}{4CF_2^2} = (x-\frac{CF_1}{2CF_2})^2$$ which gets you to $(9)$ after multiplying by $CF_2$.

Badam Baplan
  • 8,688