Question

What must be added to the following to make it a perfect square :
$\frac{3 {x}^{2} }{64} - 5 \frac{ \sqrt{21x} }{12} + \frac{157}{9}$
Options
A.
$\frac{21}{9}$
B. 1
C. 2
D.
$\frac{33}{9}$
​

Answer 1

Answer:

b)

Step-by-step explanation:

('48-$5_856336&57:67)

Answer 2

Step-by-step explanation:

for perfect square we have to transfer this qs into a^2-2ab+b^2

$\frac{3 {x}^{2} }{64} = {( \frac{ \sqrt{3}x }{8}) }^{2} = {a}^{2} \\ 5 \frac{ \sqrt{21}x }{12} = 2 \times \frac{ \sqrt{3}x }{8} \times 5 \frac{ \sqrt{7} }{3} = 2ab \\ {b}^{2} = ( {5 \frac{ \sqrt{7} }{3}})^{2} = \frac{25 \times 7}{9} = \frac{175}{9} \\ but \: we \: have \: \frac{157}{9} \\ so \: the \: number \: be \: added \: to \: make \: it \: a \: perfect \: square \: is \frac{175}{9} - \frac{157}{9} = \frac{175 - 157}{9} = \frac{18}{9} = 2$

so c is the correct answer....sorry for late reply