[Algebra 1] Need clarification in Quadratic equations

First subtract 4 from both sides to get

-(t - 7)^2 = -4

then multiply both sides by -1.
Multiply both sides by (-1) and then add 4.
OP, keep it up!

this type of self reflection on your work is what will help you become much better at math.
Going forward, these equations should always be solved by taking the square root of both sides.
In general, for a equation of the type:

a x \^ 2 + bx + c = 0

there are different ways to solve them depending on whether b and c are 0.
If b = 0, for example x \^ 2 - 9 = 0, you can just rearange it to: x \^ 2 = 9. Then take the sqrt on both sides to get x = +-3.

Now if c = 0, for example x \^ 2 - 2x = 0, you can always factor out the x and get to: x \* ( x - 2) =0. Now this product is 0 if either x = 0, so this is our first root, or if x - 2 = 0, which gives us the solutions x = 0 and x = 2.

Now the last case is if b and c are non-zero. Note that completing the square is just a way to reformulate everything using the first or second binomial formula.

For example, if we start with x\^2 + 6x + 7 = 0, we try to add some a to both sides, such that we can actually use the binomial formula. In this case, we have x\^2 + 6x. The binomial formula is (x+c)\^2 = x\^2 + 2c x + c\^2. For us the 6x corresponds to 2 c x, which implies that c = 3. So if we add and subtract c\^2 we have:

x\^2 + 6x + 3\^2 - 3 \^2 +  7 = 0,

Now the first three terms are in the correct form to use the binomial formula and we get:

(x + 3)\^2 - 9 + 7 = 0

(x + 3)\^2 - 2 = 0

And you know how to solve it from here. Note that the equation you linked is already in that form if you multiply it with -1:

(t - 7)\^2 - 4 = 0

So basically, in the example you have provided, somebody has already done the completing the square part, so there's no need for you to do that again.

0 like 0 dislike