+24 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
+129899
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
+9481 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
\(\alpha*\beta=\frac{(-b+\sqrt{b^2-4ac})(-b-\sqrt{b^2-4ac}) }{(2a)(2a)} =\frac{b^2-b\sqrt{b^2-4ac}+b\sqrt{b^2-4ac}-(b^2-4ac)}{4a^2} = \frac{4ac}{4a^2}=\frac{c}{a}\)
