Question

Help please.
Solve the equations
a) (x-2)^2 = 9(x-2)
b) x^2-4=12
c) 5x^2-5=0

Answer 1

Here a strategy is applied which is, to leave the variables and constant on opposite sides of the "=" sign, take for example the first one:
\[(x-2)^2=9(x-2)\]
This is what we could call an equation with only one variable "x", since we don't se any other letter we can only conclude it is but one variabel.
Since we also see the "=" sign, it means it is an equation and we are asked to find the exact value of the variable that satifies this equality, we can do that by applying convenient arithmetical operations in order to isolate the variable on any side of the equality sign.
And that's where your creativity come into play, how to solve this type of excercises depends solely on your knowledge and agility when applying mathematical operations and isolate the variable.
There are several methods to do this, but I'll use the notable product:
\[(a \pm b)^2 = a^2 \pm 2ab+b^2\]
This is very useful, since it allows us to expand that sum on the left side to make it more operable, and applying it:
\[(x-2)^2=9(x-2) \iff x^2-2x+4=9(x-2)\]
This last form is just the application of the notable product I stated, now let's apply distributive property to get rid of that parenthesis:
\[x^2-2x+4=9(x-2) \iff x^2-2x+4=9x-18\]
This is now more operable, and now to the main strategy is to apply the corresponding arithmetical operation to leave the numbers with a variable on one side, and those who hasn't on the other, I'll let you figure out which ones I used:
\[x^2-2x+4=9x-18 \iff x^2-2x-9x=-18-4\]
\[\iff x^2-11x=-22\]
Now, at this point you might know how an equation with the form \(ax^2+bx+c=0\)
is solved, and that's right, with the general formula, so we just accomodate this:
\[x^2-11x=-22 \iff x^2-11x+22=0 \]
And look, it is a quadratic equation which we solve with bhaskaras formula.
I must suppose you can already use this formula, I'll write it just in case and then leave the rest to you:
\[ax^2+bx+c=0 \iff x=\frac{ -b \pm \sqrt{b^2-4ac} }{ 2a }\]