Alternative method for solving quadratic formula?In the alternative method, you start with the assumptionx2+tbax+tca=0by the fundamental theorem of algebra has the form(x−r1)(x−r2)=x2−(r1+r2)x−r1r2which gives−r1−r2=ba and r1r2=ca

Resolved: "Alternative method for solving quadratic formula?In the alternative..."

e3r6a2n1dz60

2022-03-16

Alternative method for solving quadratic formula?
In the alternative method, you start with the assumption

x^{2} + t \frac{b}{a} x + t \frac{c}{a} = 0

by the fundamental theorem of algebra has the form

(x - r_{1}) (x - r_{2}) = x^{2} - (r_{1} + r_{2}) x - r_{1} r_{2}

which gives

- r_{1} - r_{2} = \frac{b}{a}

and

r_{1} r_{2} = \frac{c}{a}

sexec5m

Beginner2022-03-17Added 4 answers

Step 1
A different approach seeks to eliminate the middle term. Note if we make a substitution

u = x - d

for some constant d, we get

x = u + d

and the new equation becomes

0 = a {(u + d)}^{2} + b (u + d) + c

= a u^{2} + u [2 a d + b] + a d^{2} + b d + c

Since we want the linear term to zero out, we set

d = \frac{- b}{2 a}

and the new equation becomes

0 = a u^{2} + \frac{b^{2}}{4 a} - \frac{b^{2}}{2 a} + c

which you can simplify and solve easily.

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