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if alpha and beta are the solutions of the equation a*x(square)+b*x+c=0 show that alpha+beta=-b/a and alpha*beta=c/a

 

 

 

I have tried with 3x(square)*5x+2 but alpha*beta=c/a was wrong

bennykim0905  May 22, 2017
 #1
avatar+90052 
+2

 

Using the quadratic formula, we obtain two solutions :

 

alpha =   [ - b + √[ b^2 - 4ac] ] / {2a]       and   beta  =  [ - b - √[ b^2 - 4ac] ] / {2a]

 

So

 

alpha + beta   =      [ - b + √[ b^2 - 4ac] ] / {2a]   +   [ - b -√[ b^2 - 4ac] ] / {2a]   =  [ - 2b] / [2a]  =   -b/a

 

And  alpha * beta    =

 

[ - b + √[ b^2 - 4ac] ] / {2a]   *   [ - b + √[ b^2 - 4ac] ] / {2a]  =  [   b^2  - [b^2 - 4ac] ]  / [ 4a^2]  =

 

[ 4ac ] / [ 4a^2]  =   c / a

 

 

 

 

cool cool cool

CPhill  May 22, 2017
 #2
avatar+7324 
+1

Here, the solutions are the x values that make this equation true.

ax2 + bx + c = 0

 

We can use the quadratic formula to solve for x.

x = \({-b \pm \sqrt{b^2-4ac} \over 2a}\)

 

So..the two solutions are

\(\alpha=\frac{-b+\sqrt{b^2-4ac}}{2a} \quad \text{and} \quad \beta=\frac{-b-\sqrt{b^2-4ac}}{2a} \)

 

And...

\(\alpha+\beta=\frac{-b+\sqrt{b^2-4ac}+-b-\sqrt{b^2-4ac}}{2a} = \frac{-2b}{2a}=\frac{-b}{a}\)

 

and

\(\alpha*\beta=\frac{(-b+\sqrt{b^2-4ac})(-b-\sqrt{b^2-4ac}) }{(2a)(2a)} =\frac{b^2-b\sqrt{b^2-4ac}+b\sqrt{b^2-4ac}-(b^2-4ac)}{4a^2} = \frac{4ac}{4a^2}=\frac{c}{a}\)

hectictar  May 22, 2017
edited by hectictar  May 22, 2017

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