Question

The arrangement of following quadratic equations in the ascending order of their product of roots.
A: $3x^2-5x+7=0$
B: $\sqrt{3}{x}^2-10x-8\sqrt{3}=0$
C: $6x^2-5x+1=0$
D: $x^2-5x+6=0$

• A. $B,C,A,D$
• B. $C,A,B,D$
• C. $C,D,B,A$
• D. $D,A,B,C$

Answer 1

The correct option is A $B,C,A,D$

$ForA:$

$2x^2-5x+7=0$

$2x^2-7x+2x+7=0$

$x(2x-7)-1(2x-7)=0$

$(2x-7)(x-1)=0$

$\therefore$ $x=\frac{7}{2}$ and $x=1$

$\Rightarrow$ Product of the roots $=\frac{7}{2}\times 1=\frac{7}{2}=3.5$ -------- ( 1 )

$ForB:$

$\sqrt{3}{x}^2-10x-8\sqrt{3}=0$

$\sqrt{3}{x}^2-12x+2x-8\sqrt{3}=0$

$\sqrt{3}x(x-4\sqrt{3})+2(x-4\sqrt{3})=0$

$(x-4\sqrt{3})(\sqrt{3}x+2)=0$

$\therefore$ $x=4\sqrt{3}$ and $x=\frac{-2}{\sqrt{3}}$

$\Rightarrow$ Product of the roots $=4\sqrt{3}\times \left(\frac{-2}{\sqrt{3}}\right)=-8$ --------- ( 2 )

$ForC:$

$6x^2-5x+1=0$

$6x^2-6x+x+1=0$

$6x(x-1)-1(x-1)=0$

$(x-1)(6x-1)=0$

$\therefore$ $x=1$ and $x=\frac{1}{6}$

$\Rightarrow$ Product of roots $=1\times \frac{1}{6}=\frac{1}{6}=0.17$ ---------- ( 3 )

$ForD:$

$x^2-5x+6=0$

$x^2-6x+x+6=0$

$x(x-6)-1(x-6)=0$

$(x-6)(x-1)=0$

$\therefore$ $x=6$ and $x=1$

$\Rightarrow$ Product of roots $=6\times 1=6$ ---------- ( 4 )

From ( 1 ), ( 2 ), ( 3 ) and ( 4 ) ascending order of product of roots are $B,C,A,D$