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# question

+4
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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
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#1
+79894
+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

CPhill  May 22, 2017
#2
+5589
+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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