Question 261529
{{{z - 7z^(1/2) + 12 = 0}}}
The key to solving this equation is to recognize that {{{z = (z^(1/2))^2}}}. This is what makes this a quadratic equation. If you have trouble seeing this, then we can use a temporary variable:
Let {{{q = z^(1/2)}}}
then {{{q^2 = z}}}
Replacing z and {{{sqrt(z)}}} with {{{q^2}}} and q respectively your equation becomes:
{{{q^2 -7q + 12 = 0}}}
This is clearly a quadratic equation. We can solve this by factoring:
{{{(q-3)(q-4) = 0}}}
From the Zero Product Property we know that this product can be zero only if one of the factors is zero. So:
{{{q-3 = 0}}} or {{{q-4 = 0}}}
Solving these we get:
{{{q = 3}}} or {{{q = 4}}}<br>
Of course we are not interested in what q is. We want a solution for z. So at this point we replace q with {{{sqrt(z)}}}:
{{{z^(1/2) = 3}}} or {{{z^(1/2) = 4}}}
We now have one more step. We need to square both sides of these equations giving us:
{{{z = 9}}} or {{{z = 16}}}<br>
Using a temporary variable is not required. But they can be helpful until you get used to working without them. Here is a solution which does not use a temporary variable:
{{{z - 7z^(1/2) + 12 = 0}}}
Factor:
{{{(z^(1/2) - 3)(z^(1/2) - 4) = 0}}}
Zero Product Property:
{{{z^(1/2) - 3 = 0}}} or {{{z^(1/2) - 4 = 0}}}
Solve:
{{{z^(1/2) = 3}}} or {{{z^(1/2) = 4}}}
{{{z = 9}}} or {{{z = 16}}}