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Answer :
We are given the equation
[tex]$$
h^2 = 169.
$$[/tex]
To solve for [tex]$h$[/tex], we take the square root of both sides. Remember that if
[tex]$$
x^2 = a,
$$[/tex]
then
[tex]$$
x = \sqrt{a} \quad \text{or} \quad x = -\sqrt{a}.
$$[/tex]
In our case, taking the square root of both sides gives:
[tex]$$
h = \sqrt{169} \quad \text{or} \quad h = -\sqrt{169}.
$$[/tex]
Since the square root of [tex]$169$[/tex] is [tex]$13$[/tex], the two solutions are:
[tex]$$
h = 13 \quad \text{or} \quad h = -13.
$$[/tex]
Thus, the values of [tex]$h$[/tex] that satisfy the equation are
[tex]$$
\boxed{13 \text{ and } -13}.
$$[/tex]
[tex]$$
h^2 = 169.
$$[/tex]
To solve for [tex]$h$[/tex], we take the square root of both sides. Remember that if
[tex]$$
x^2 = a,
$$[/tex]
then
[tex]$$
x = \sqrt{a} \quad \text{or} \quad x = -\sqrt{a}.
$$[/tex]
In our case, taking the square root of both sides gives:
[tex]$$
h = \sqrt{169} \quad \text{or} \quad h = -\sqrt{169}.
$$[/tex]
Since the square root of [tex]$169$[/tex] is [tex]$13$[/tex], the two solutions are:
[tex]$$
h = 13 \quad \text{or} \quad h = -13.
$$[/tex]
Thus, the values of [tex]$h$[/tex] that satisfy the equation are
[tex]$$
\boxed{13 \text{ and } -13}.
$$[/tex]
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