How to Turn #ax^2+bx+c=0=>Ax+B=0#?
Substitute #x^2=y#
#ay+bsqrt(y)+c=0#
If #bsqrt(y)=0# ,then
#ay+c=0#
so #ay+c=+-bsqrt(y)#
if #y=z-c/(4a)# then
#az+(c-c/4)=+-bsqrt(z-c/(4a))#
#Az+B=Cw#
#w=A/Cz+B/C=>Pz+Q#
Substitute
If
so
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1 Answer
A few thoughts...
Explanation:
Given any polynomial in the form:
#a_n x^n + a_(n-1) x^(n-1) + ... + a_0" "# with#a_n != 0#
we can get a simplified 'depressed' form using a linear substitution:
#t = x+a_(n-1)/(n a_n)#
This will give us a polynomial in the form:
#b_n t^n + b_(n-2) t^(n-2) + b_(n-3) t_(n-3) + ... + b_0#
with no term in
In the case of a quadratic, we find:
#ax^2+bx+c = a(x+b/(2a))^2+(c-b^2/(4a)) = at^2+(c-b^2/(4a))#
where
which then makes the finding of zeros just a matter of taking the square root then reversing the linear transformation to find:
#x = (-b+-sqrt(b^2-4ac))/(2a)#
Of course, you could apply further substitutions:
#{ (A = a), (B = c-b^2/a), (y = t^2) :}#
to get:
#ax^2+bx+c = Ay+B#
thus transforming
That just gives you a few more steps to reverse.
