Question

If rational numbers $a,b,c,d$ are in G.P., then roots of equation $(a-c{)}^2{x}^2+(b-c{)}^{2}x+(b-d{)}^2=(a-d{)}^2$ are necessarily

• A. Imaginary
• B. Irrational
• C. rational
• D. real and distinct

Answer 1

The correct option is C rational

Given $a,b,c,d$ are in G.P. $\therefore \frac{b}{a}=\frac{c}{b}=\frac{d}{c}$ or
$b^2=ac,c^2=bd,bc=ad...\left(1\right)$

If sum of the coefficients of $A{X}^2+BX+K$ is zero i.e. $A+B+K=0$ then one of the root is $1$, which is rational and a,b,c,d are rational

So $(a-c{)}^2{x}^2+(b-c{)}^{2}x+(b-d{)}^2=(a-d{)}^2$are rationals means sum of the coefficients of $A{X}^2+BX+K=0$ are rational therefore other root must be rational which means both roots are rational Explanation: Sum of the coefficients of quadratic equation $(a-c{)}^2{x}^2+(b-c{)}^{2}x+(b-d{)}^2=$

$(a-d{)}^2=a^2+c^2-2ac+b^2+c^2-2bc+b^2+d^2-2bd-a^2-d^2+2ad$
$2b^2-2ac+2c^2-2bd-2bc+2ad=0+0+0=(from...(1\left)\right)=0$