For ax2 + bx + c = 0 with roots r1, r2, state Vieta’s relations.
Solution
Correct Answer: Option B
For the quadratic equation \(ax^2+bx+c=0\) (with \(a\neq0\)) let the roots be \(r_1\) and \(r_2\). Then the polynomial can be written in factored form
\[
a(x-r_1)(x-r_2)=ax^2-a(r_1+r_2)x+a\,r_1r_2.
\]
Comparing coefficients with \(ax^2+bx+c\) gives
\[
-a(r_1+r_2)=b\quad\Rightarrow\quad r_1+r_2=-\frac{b}{a},
\]
\[
a\,r_1r_2=c\quad\Rightarrow\quad r_1r_2=\frac{c}{a}.
\]
Thus Vieta’s relations are \(r_1+r_2=-\dfrac{b}{a}\) and \(r_1r_2=\dfrac{c}{a}\). (If \(a=1\) these reduce to \(r_1+r_2=-b\) and \(r_1r_2=c\).)
\[
a(x-r_1)(x-r_2)=ax^2-a(r_1+r_2)x+a\,r_1r_2.
\]
Comparing coefficients with \(ax^2+bx+c\) gives
\[
-a(r_1+r_2)=b\quad\Rightarrow\quad r_1+r_2=-\frac{b}{a},
\]
\[
a\,r_1r_2=c\quad\Rightarrow\quad r_1r_2=\frac{c}{a}.
\]
Thus Vieta’s relations are \(r_1+r_2=-\dfrac{b}{a}\) and \(r_1r_2=\dfrac{c}{a}\). (If \(a=1\) these reduce to \(r_1+r_2=-b\) and \(r_1r_2=c\).)
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