Solving for x in terms of y: Methods and Alternatives

  • Thread starter phymatter
  • Start date
In summary, the conversation was about a method for solving equations using completing the square. The quadratic formula is derived from this method. The process involves adding a term to both sides of the equation and then using square roots to solve for x. The attachment involves solving an equation using this method.
  • #1
phymatter
131
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Can Anyone Understand this?

See the attachment///
Can Anyone Understand this?
I mean what metod has he followed to solve this equation for x in terms of y , does someone have any method to solve it other than the quadratic formula?/
 

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  • #2


It's called "completing the square", and it's where the quadratic formula comes from. Remember, if I have:
[tex]x^2+2ax + a^2 = 0[/tex] , I can write it as
[tex](x+a)^2 = 0[/tex] .
So if I have
[tex]ax^2+bx+c = 0[/tex]
I can write it as:
[tex]x^2+\frac{b}{a}x+(\frac{b}{2a})^2 =(\frac{b}{2a})^2-\frac{c}{a}[/tex]
where I have added [tex](\frac{b}{2a})^2[/tex] to both sides. This is also:
[tex](x+\frac{b}{2a})^2 = (\frac{b}{2a})^2-\frac{c}{a}[/tex]
or:
[tex](x+\frac{b}{2a}) =\pm\sqrt{(\frac{b}{2a})^2-\frac{c}{a}}[/tex]
or:
[tex]x =-\frac{b}{2a}\pm\sqrt{(\frac{b}{2a})^2-\frac{c}{a}}[/tex]
or:
[tex]x =\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]

This is what they have done, but with [tex]\frac{7y+13}{24}[/tex] playing the role of [tex]\frac{b}{2a}[/tex]
 
  • #3


Thanks friend
 

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