To solve for a variable, you need to get it (x) by itself.
x² - 14 + 31 = 63 Add 31 to -14
x² + 17 = 63 Subtract 17 from both sides of the equation
x² = 46 Square root both sides
x =

Check your work by plugging in

for x and then

.

² - 14 + 31 = 63 Square

46 - 14 + 31 = 63 Subtract 14 from 46
32 + 31 = 63 Add
63 = 63

² - 14 + 31 = 63 Square

46 - 14 + 31 = 63 Subtract 14 from 46
32 + 31 = 63 Add
63 = 63
So,
x is
and 
.
if any concerns or questions on solution, just comment
Answer:
Polygon q’s area is one fourth of polygon p’s area
Step-by-step explanation:
we know that
If two figures are similar, then the ratio of its areas is equal to the scale factor squared
Let
z-----> the scale factor
x-----> polygon q’s area
y-----> polygon p’s area
so

In this problem we have

substitute



therefore
Polygon q’s area is one fourth of polygon p’s area
Point, line and plane are undefined termseverything else is defined based on them - there are no other undefined terms in euclid's geometry - only those 3 listed above are undefined terms