Question

If $a,b,c,d$ and $p$ are distinct non-zero real numbers such that $(a^2+b^2+c^2)p^2-2(ab+bc+dc)p+(b^2+c^2+d^2)\le 0$, then $ac$ is equal to

• A. $b^2$
• B. $d^2$
• C. $p^2$
• D. $b^2+d^2$

Answer 1

The correct option is A $b^2$

$(a^2+b^2+c^2)p^2-2p(ab+bc+cd)+(b^2+c^2+d^2)\le 0$
$\Rightarrow (ap-b{)}^2+(bp-c{)}^2+(cp-d{)}^2\le 0$
But $(ap-b{)}^2+(bp-c{)}^2+(cp-d{)}^2\ge 0$
$\therefore ap-b=0;bp-c=0;cp-d=0$
$\Rightarrow p=\frac{b}{a}=\frac{c}{b}=\frac{d}{c}$
$\Rightarrow a=\frac{b}{p},c=bp$
$\therefore ac=b^2$