Thread: solution of the quadratic equation

Question



Attempt

Using the Formula I got 4

Is my answer correct?

Originally Posted by mj.alawami

[SIZE=4]Question

From what you learned about factoring quadratics and special factoring formulas, you know that the above can be restated as:

. . . . .

What result do you get when you solve the factors?

Originally Posted by mj.alawami

Using the Formula I got 4

Is my answer correct?

The Quadratic Formula gives a result (using a = 1, b = 0, and c = -4) that agrees with the "by factoring" result. What formula did you use?

Note: The answer to any "solving" problem can be checked by plugging it back into the original exercise. In your case:

. . . . .

Originally Posted by stapel

From what you learned about factoring quadratics and special factoring formulas, you know that the above can be restated as:

. . . . .

What result do you get when you solve the factors?


The Quadratic Formula gives a result (using a = 1, b = 0, and c = -4) that agrees with the "by factoring" result. What formula did you use?

Note: The answer to any "solving" problem can be checked by plugging it back into the original exercise. In your case:

. . . . .

I have done

Then I Used the Quadratic Formula and Got 4

Originally Posted by mj.alawami

I have done

Then I Used the Quadratic Formula and Got 4

So to Finalize the answer is -2

Originally Posted by mj.alawami

So to Finalize the answer is -2

no

Originally Posted by mj.alawami

I have done

Then I Used the Quadratic Formula and Got 4

and the purpose of adding zero to the left side was...?

-4 is your constant term, that is the "c" in your formula. thus you have a = 1, b = 0 and c = -4

applying the quadratic formula, we have




for problems like these, the quadratic formula is overkill. either do what stapel did, or just add 4 to both sides and then square root both sides (and remember that the right side will have a +/- with it, since there are two square roots to a number)