Why isn't y = a ( x 2 − S x + P ) same with x 2 − S x + PIf we have roots of the function y = a x 2 + b x + c we can calculate S = − b a and also P = c a . Then we know that we can form the function this way: x 2 − S x + PSo on the other side we know that we have the function f ( x ) = y in different ways: y = a x 2 + b x + c( α and β are roots of the quadratic function) y = a ( x − α ) ( x − β )And my question is here: y = a ( x 2 − S x + P )Actually know that how we can form the qudratic equation using x 2 − S x + P, but the function must be like y = a ( x 2 − S x + P ). Actually I don't know that why we add a. I know it will be removed when ( a ) ( − b a ). But I don't know that what is y = a ( x 2 − S x + P ) different whitout a!

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Why isn&#039;t   y  =  a  (      x    2    −  S  x  +  P  ) same with       x    2    −  S  x  +  PIf we have roots of the function   y  =  a      x    2    +  b  x  +  c we can calculate   S  =            −      b        a   and also   P  =      c    a  . Then we know that we can form the function this way:      x    2    −  S  x  +  PSo on the other side we know that we have the function   f  (  x  )  =  y in different ways:  y  =  a      x    2    +  b  x  +  c(  α and   β are roots of the quadratic function)  y  =  a  (  x  −  α  )  (  x  −  β  )And my question is here:  y  =  a  (      x    2    −  S  x  +  P  )Actually know that how we can form the qudratic equation using       x    2    −  S  x  +  P, but the function must be like   y  =  a  (      x    2    −  S  x  +  P  ). Actually I don&#039;t know that why we add a. I know it will be removed when   (  a  )  (            −      b        a    ). But I don&#039;t know that what is   y  =  a  (      x    2    −  S  x  +  P  ) different whitout a!

Dana Simmons · Accepted Answer

Step 1a is there just to include the possibility of a quadratic having a coefficient other than 1 for       2          n      d       degree term       x    2  .Otherwise, how will you make a quadratic of the form:  y  =  (  a      x    2    +  b  x  +  c  )(where   a  ≠  1)merely by taking it like:  y  =  (  x  −  α  )  (  x  −  β  )  ?Step 2Clearly, coefficient of       x    2   is 1 in this.So, while assuming a quadratic when its roots are known, we take it  y  =  a  (  x  −  α  )  (  x  −  β  )  ,just to be on the safer side.

Christopher Saunders · Accepted Answer

Step 1Two quadratic equations      y    1    =  (  x  −  b  )  (  x  −  c  )and      y    2    =  a  (  x  −  b  )  (  x  −  c  )have the same roots but not the same values at any other points. They are different function which share common roots.For example      y    1    =      x    2    +  5  x  +  6and      y    2    =  3      x    2    +  15  x  +  18are different functions with the same roots.Step 2Notice that      y    1    (  2  )  =  20while      y    2    (  2  )  =  60What you like to say is that the two quadratic equations,      x    2    +  5  x  +  6  =  0and  3      x    2    +  15  x  +  18  =  0are equivalent because we can factor 3 out and 3 is not zero so it does not change the roots.

Why isn't y=a(x^2-Sx+P) same with x^2-Sx+P

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