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Author Topic: [Solutions] All Mathematics Answers To The Quiz  (Read 50837 times)

Offline SLNAgency

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[Solutions] All Mathematics Answers To The Quiz
on: March 04, 2020, 03:04:43 PM


[Mathematics 3] Solving Quadratic Equations Solutions

A quadratic equation is an equation that could be written as ax2 + bx + c = 0; when a 0. There are three basic methods for solving quadratic equations:

I. Factoring
II. Using the quadratic formula, and
III. Completing the square.


1. Solve this equation x2 − x − 12 = 0
(x − 4)(x + 3) = 0
Now at this point we’ve got a product of two terms that is equal to zero. This means that at least one of the following must be true.
x − 4 = 0 OR x + 3 = 0
x = 4 OR −3

2. Solve ⇒ x2 + 40 + 14x = 0
   (x + 4)(x + 10) = 0
   Now, we once again have a product of two terms that equals zero so we know that one or both of them have to be zero. So, technically we need to set each one equal to zero and solve. However, this is usually easy enough to do in our heads and so from now on we will be doing this solving in our head.

The solutions to this equation are, x = −4 AND x = −10



3. Solve ⇒ y2 + 12y + 36 = 0
   (y + 6)2 = 0
   (y + 6)(y + 6) = 0

   in this case we’ve got a perfect square. We broke up the square to denote that we really do have an application of the zero factor property. However, we usually don’t do that. We usually will go straight to the answer from the squared part.

The solution to the equation in this case is, y = −6

4. Solve 5x3 − 5x2 − 10x = 0
The first thing to do is factor this equation as much as possible. In this case that means factoring out the greatest common factor first. Here is the factored form of this equation.
   
   5x(x2 − x − 2) = 0
   5x(x − 2)(x + 1) = 0

   Now, the zero factor property will still hold here. In this case we have a product of three terms that is zero. The only way this product can be zero is if one of the terms is zero. This means that,

   5x = 0 ⇒ x = 0
   x − 2 = 0 ⇒ x = 2
   x + 1 = 0 ⇒ x = −1
   x = 2 OR -1

5. Solve ⇒ 10z2 + 19z + 6 = 0
   ⇒ (5z + 2)(2z +3) = 0
   ⇒ z = −2/5 and z = −3/2



6. Solve ⇒ 3x2 − 2x − 8 = 0
   ⇒ (3x + 4)(x − 2) = 0
   ⇒ x = −4/3 and x = 2

7. Solve ⇒ 5x2 − 2x = 0
 
Now, notice that all we can do for factoring is to factor an x
 out of everything. Doing this gives,
    x(5x − 2) = 0
    x = 0 OR (5x − 2) = 0
    x = 0 OR 5x = 2
    x = 0 OR 2/5

8. Solve the equation 25y2 = 3
    y2 = 3/25
    ⇒ y = ±√3/25
     y = ±√3/5

9. Solve ⇒ x2 = 100
   x = ±√100
   x = ±10

10. Solve ⇒ 4z2 = −49
   z2 = −49/4
   ⇒ z = ±√−49/4
    z = ±√−7/2




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