Quadratic Trinomials, Quadratic Equations, and the Quadratic Formula Practice Problems

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Quadratic Trinomials, Equations, and Formula Practice Problems

Directions: Use scratch paper to solve the following problem. You can check your answer at the end of this chapter.

Practice

1. Factor z² – 6z + 9.
2. Solve 4x² = 100.
3. If (x – 8)(x + 5) = 0, what are the two possible values of x?
4. If x² – 8x = 20, which of the following could be a value of x² + 8x? –20, 28, 108, or 180
5. 

What is the equation represented in the graph?

1. y = x² + 3
2. y = x² – 3
3. y = (x + 3)²
4. y = (x – 3)²
5. y = (x – 1)³

Solutions

1. To find the factors, follow the FOIL method in reverse:

   z² – 6z + 9

   The product of the last pair of terms equals + 9. There are a few possibilities for these terms: 3 and 3 (because 3 × 3 = + 9), –3 and –3 (because –3 × –3 = + 9), 9 and 1 (because 9 × 1 = + 9), –9 and –1 (because –9 × –1 = + 9).

   The sum of the products of the outer pair of terms and the inner pair of terms equals –6z. So we must choose the two last terms from the list of possibilities that would add up to –6. The only possibility is –3 and –3. Therefore, we know that the last terms are –3 and –3.

   The product of the first pair of terms equals z². The most likely two terms for the first pair is z and z because z × z = z² .

   Therefore, the factors are (z – 3)(z – 3).

2. Make one side of the equation zero. Subtract 100 from both sides of the equation: 4x² – 100 = 100 – 100. Now, simplify: 4x² – 100 = 0.

   Factor out the greatest common factor: 4(x² – 25) = 0. Factor using the difference of two squares: 4(x – 5)(x + 5) = 0. Divide both sides of the equation by 4: . Time to simplify again: (x – 5)(x + 5) = 0. Set each factor equal to zero: x – 5 = 0 and x + 5 = 0. Solve the equations: x = 5 and x = –5. The solutions for the quadratic equation are 5 and –5.

3. If (x – 8)(x + 5) = 0, then one (or both) of the factors must equal 0.

   x – 8 = 0 if x = 8 because 8 – 8 = 0.

   x + 5 = 0 if x = –5 because –5 + 5 = 0.

   So, the two values of x that make (x – 8)(x + 5) = 0 are x = 8 and x = –5.

4. This question requires several steps to answer. First, you must determine the possible values of x considering that x² – 8x = 20. To find the possible x-values, rewrite x² – 8x = 20 as x² – 8x – 20 = 0, factor, and then use the zero product rule; x² – 8x – 20 = 0 is factored as (x – 10)(x + 2). So, possible values of x are x = 10 and x = –2 because 10 – 10 = 0 and –2 + 2 = 0.

   Now, to find possible values of x2 + 8x, plug in the x-values. If x = –2, then x² + 8x = (–2)2 + (8)(–2) = 4 + (–16) = –12. None of the answer choices are –12, so try x = 10. If x = 10, then x² + 8x = 102 + (8)(10) = 100 + 80 = 180.

5. This graph is identical to a graph of y = x² except it is moved down 3 so that it intersects the y-axis at –3 instead of 0. Each y-value is 3 less than the corresponding y-value in y = x², so its equation is therefore y = x² – 3 (choice b).