I. quad p^{2}-18 p+77=0
II. quad 3 q^{2}-25 | vedclass

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(C) Quantity $1: p^{2}-18 p+77=0$
$\Rightarrow p^{2}-11 p-7 p+77=0$
$\Rightarrow (p-11)(p-7)=0$
$\Rightarrow p = 11, 7$
Quantity $2: 3 q^{2}-25 q+

Vedclass · Accepted Answer

Quantity $I \geq$ Quantity $II$

Vedclass · Answer

(C) Quantity $1: p^{2}-18 p+77=0$
$\Rightarrow p^{2}-11 p-7 p+77=0$
$\Rightarrow (p-11)(p-7)=0$
$\Rightarrow p = 11, 7$
Quantity $2: 3 q^{2}-25 q+28=0$
$\Rightarrow 3 q^{2}-21 q-4 q+28=0$
$\Rightarrow 3 q(q-7)-4(q-7)=0$
$\Rightarrow (3 q-4)(q-7)=0$
$\Rightarrow q = 7, \frac{4}{3} \approx 1.33$
Comparing the values: The set of values for $p$ is ${7, 11}$ and for $q$ is ${1.33, 7}$.
Since all values of $p$ are greater than or equal to the values of $q$,we conclude that Quantity $1 \geq$ Quantity $2$.

$I. \quad p^{2}-18 p+77=0$
$II. \quad 3 q^{2}-25 q+28=0$
$Quantity \, 1$: Value of $p$
$Quantity \, 2$: Value of $q$

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$I. \quad p^{2}-18 p+77=0$ $II. \quad 3 q^{2}-25 q+28=0$ $Quantity \, 1$: Value of $p$ $Quantity \, 2$: Value of $q$