Subject

    গণিত

    Topic

    Algebra

    For ax2+ bx + c = 0 with roots r1, r2, state Vieta’s relations.

    ক)
    r1 + r2 = b/a, r1 r2 = −c/a
    খ)
    r1 + r2 = −b/a, r1 r2 = c/a
    গ)
    r1 + r2 = −c/a, r1 r2 = b/a
    ঘ)
    r1 + r2 = c/a, r1 r2 = −b/a

    Explanation

    For the quadratic equation ax2+bx+c=0ax^2+bx+c=0 (with a≠0a\neq0) let the roots be r1r_1 and r2r_2. Then the polynomial can be written in factored form a(x−r1)(x−r2)=ax2−a(r1+r2)x+a r1r2. a(x-r_1)(x-r_2)=ax^2-a(r_1+r_2)x+a\,r_1r_2. Comparing coefficients with ax2+bx+cax^2+bx+c gives −a(r1+r2)=b⇒r1+r2=−ba, -a(r_1+r_2)=b\quad\Rightarrow\quad r_1+r_2=-\frac{b}{a}, a r1r2=c⇒r1r2=ca. a\,r_1r_2=c\quad\Rightarrow\quad r_1r_2=\frac{c}{a}. Thus Vieta’s relations are r1+r2=−bar_1+r_2=-\dfrac{b}{a} and r1r2=car_1r_2=\dfrac{c}{a}. (If a=1a=1 these reduce to r1+r2=−br_1+r_2=-b and r1r2=cr_1r_2=c.)

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