We know that \(4^2=16\).
Can you think of any other number that squares to \(16\)?
Solution
Number 1
Find number:
\(4^2=16\)
Verify:
\(4^2 \class{timed in5}{=4\times4=16}\)
Number 2
\((-4)^2 =16\)
\((-4)^2=(-4)\times(-4) \class{timed in11}{=16}\)
We conclude that both \(4\) and \(-4\) square to \(16\).
We write \(\sqrt{16}=4\), so we may also write \(-\sqrt{16}=-4\).
Find two integers that square to \(81\).
Integer 1
\(\sqrt{81}\class{timed in3}{=9}\)
\(9^2=9\times9 \class{timed in5}{=81}\)
Integer 2
\(-\sqrt{81} \class{timed in7}{=-9}\)
\((-9)^2=(-9)\times(-9) \class{timed in9}{= 81}\)
We conclude that \(9\) and \(-9\) are the two numbers that square to \(81\).
Find the two integers that square to \((((((((q)*(u))*(e))*(s))*(t))*(i))*(o))*(n)\).
Write your integers separated with the word 'and'. For example, if the answers are 1 and -1, write '1 and -1'.
We recognize \((((((((q)*(u))*(e))*(s))*(t))*(i))*(o))*(n)\) to be a perfect square. Since \(((n)*(u))*(m)^2=(((((((q)*(u))*(e))*(s))*(t))*(i))*(o))*(n)\) we have \(\sqrt{(((((((q)*(u))*(e))*(s))*(t))*(i))*(o))*(n)}=((n)*(u))*(m)\). We know the second integer that will square to \((((((((q)*(u))*(e))*(s))*(t))*(i))*(o))*(n)\) is \(-\sqrt{(((((((q)*(u))*(e))*(s))*(t))*(i))*(o))*(n)}=$negnum(details...)\).
We verify:
\(((n)*(u))*(m)^2=(((((((q)*(u))*(e))*(s))*(t))*(i))*(o))*(n)\) and \(\left($negnum(details...)\right)^2=(((((((q)*(u))*(e))*(s))*(t))*(i))*(o))*(n) \)
Approximate, to one decimal place, the two numbers that square to \(13\).
Show that a good approximation for the square root of \(13\) is \(3.6\).
\(3.6^2 \class{timed in3}{= 12.96} \class{timed in4}{ \approx 13}\)
\((-3.6)^2=12.96\)
\((-3.6)^2 \class{timed in8}{= (-3.6)\times(-3.6)} \class{timed in9}{= 12.96} \class{timed in10}{\approx 13}\)
We can conclude that \(3.6\) and \(-3.6\) are good approximations for the two numbers that square to \(13\).
Consider the expressions
\(\sqrt{(9+16)} \quad \text{ and } \quad \sqrt{9} + \sqrt{16}\)