Question 347066
{{{(x-5)^2-4(x-5)-21=0}}}
There's a clever way and a long way to solve this equation. I'll do both.<br>
Clever way, using a temporary variable. This involves noticing that this equation is quadratic in (x-5). It might help you to see this if I use a temporary variable. Let q = x-5. Substituting q for x-5, your equation becomes:
{{{q^2 -4q -21 = 0}}}
This equation is easily solved for q. Just factor it:
{{{(q-7)(q+3) = 0}}}
Use the Zero Product property:
q-7 = 0 or q+3 = 0
Solving these we get:
q = 7 or q = -3
And last we replace q with x-5:
x-5 = 7 or x-5 = -3
Solving these we get:
x = 12 or x = 2<br>
Clever way without a temporary variable. Eventually you will be able to solve this kind of problem without a temporary variable:
{{{(x-5)^2-4(x-5)-21=0}}}
{{{((x-5)-7)((x-5)+3) = 0}}}
{{{(x-12)(x-2) = 0}}}
x-12 = 0 or x-2 = 0
x = 12 or x = 2
Without the temporary variable this is a very short solution.<br>
The long way to do this problem is to start by simplifying the equation. We'll square (x-5) first:
{{{(x-5)^2-4(x-5)-21=0}}}
{{{x^2 -10x +25-4(x-5)-21=0}}}
Multiply next:
{{{x^2 -10x +25-4x +20-21=0}}}
Combine like terms:
{{{x^2 -14x +24=0}}}
Factor:
{{{(x-12)(x-2) = 0}}}
Zero Product property:
x-12 = 0 or x-2 = 0
Solve:
x = 12 or x = 2