prove-that-x-1-x-2-b-a-x-1-and-x-2-are-roots-of-ax-2-bx-c-0-

Question Number 132475 by Study last updated on 14/Feb/21

p r o v e t h a t x_{1} + x_{2} = - \frac{b}{a}

x_{1} a n d x_{2} a r e r o o t s o f {a x}^{2} + b x + c = 0

Commented by MJS_new last updated on 14/Feb/21

{it}^{'} s just the solution formula again . solve

the quadratic and add the solutions . {where}^{'} s

the problem ?

Commented by EDWIN88 last updated on 14/Feb/21

may be the problem why x_{1} + x_{2} = - \frac{b}{a}

Commented by MJS_new last updated on 14/Feb/21

as I said, solve by using the usual formula

Commented by EDWIN88 last updated on 14/Feb/21

yes

Answered by rs4089 last updated on 14/Feb/21

l e t x_{1} a n d x_{2} a r e r o o t s o f {e q}^{n} {a x}^{2} + b x + c = 0

w e c a n w r i t e n a s {a x}^{2} + b x + c = a (x - x_{1}) (x - x_{1})

c o m p a r e c o f f i c i e n t o f x b o t h s i d e \dots

b = - a (x_{1} + x_{2})

s o x_{1} + x_{2} = \frac{- b}{a}

Answered by physicstutes last updated on 14/Feb/21

Or begin with the general solution to the quadratic :

x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}

say, x_{1} = \frac{- b + \sqrt{b^{2} - 4 a c}}{2 a} and x_{2} = \frac{- b - \sqrt{b^{2} - 4 a c}}{2 a}

x_{1} + x_{2} = \frac{- b + \sqrt{b^{2} - 4 a c}}{2 a} + \frac{- b - \sqrt{b^{2} - 2 a c}}{2 a} = \frac{- b + \sqrt{b^{2} - 4 a 4} - b - \sqrt{b^{2} - 4 a c}}{2 a}

x_{1} + x_{2} = \frac{- b - b}{2 a} = \frac{- b}{a}

hence x_{1} + x_{2} = - \frac{b}{a}

similarly it can be shown that x_{1} x_{2} = \frac{c}{a}

Commented by MJS_new last updated on 14/Feb/21

{that}^{'} s exactly what I recommended

if we answer the easiest questions, people

wil never learn anything . you can do all

their thinking - at least all their homework -

for them for all eternity .

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