Quadratic practice sheet - All this - Free Printable

Let's solve each quadratic equation step by step. We'll use factoring, completing the square, or the quadratic formula as appropriate.

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1. $ 2x^2 - 2x - 24 = 0 $

Step 1: Factor out the GCF if possible.
Here, we can factor out a 2:

$$
2(x^2 - x - 12) = 0
$$

Now factor the quadratic inside:

$$
x^2 - x - 12 = (x - 4)(x + 3)
$$

So:
$$
2(x - 4)(x + 3) = 0
$$

Set each factor equal to zero:

- $ x - 4 = 0 $ → $ x = 4 $
- $ x + 3 = 0 $ → $ x = -3 $

✔ Solutions: $ x = 4 $, $ x = -3 $

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2. $ x^2 + 10x + 16 = 0 $

We look for two numbers that multiply to 16 and add to 10.

Try: $ 8 \times 2 = 16 $, $ 8 + 2 = 10 $ → Yes!

$$
x^2 + 10x + 16 = (x + 8)(x + 2) = 0
$$

Set each factor to zero:

- $ x + 8 = 0 $ → $ x = -8 $
- $ x + 2 = 0 $ → $ x = -2 $

✔ Solutions: $ x = -8 $, $ x = -2 $

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3. $ 4x^2 + 15x + 9 = 0 $

We'll try factoring. Look for two numbers that multiply to $ 4 \times 9 = 36 $ and add to 15.

Try: $ 12 $ and $ 3 $ → $ 12 \times 3 = 36 $, $ 12 + 3 = 15 $ → Yes!

Now split the middle term:

$$
4x^2 + 12x + 3x + 9 = 0
$$

Group:

$$
(4x^2 + 12x) + (3x + 9) = 4x(x + 3) + 3(x + 3) = (4x + 3)(x + 3)
$$

So:

$$
(4x + 3)(x + 3) = 0
$$

Set each factor to zero:

- $ 4x + 3 = 0 $ → $ x = -\frac{3}{4} $
- $ x + 3 = 0 $ → $ x = -3 $

✔ Solutions: $ x = -\frac{3}{4} $, $ x = -3 $

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4. $ x^2 - 13x + 42 = 0 $

Find two numbers that multiply to 42 and add to -13.

Try: $ -6 $ and $ -7 $ → $ (-6)(-7) = 42 $, $ -6 + (-7) = -13 $ → Yes!

$$
x^2 - 13x + 42 = (x - 6)(x - 7) = 0
$$

Set each factor to zero:

- $ x - 6 = 0 $ → $ x = 6 $
- $ x - 7 = 0 $ → $ x = 7 $

✔ Solutions: $ x = 6 $, $ x = 7 $

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5. $ 6x^2 - x - 40 = 0 $

Use the AC method: $ a = 6 $, $ b = -1 $, $ c = -40 $

$ ac = 6 \times (-40) = -240 $

Find two numbers that multiply to -240 and add to -1.

Try: $ -16 $ and $ 15 $ → $ -16 \times 15 = -240 $, $ -16 + 15 = -1 $ → Yes!

Split the middle term:

$$
6x^2 - 16x + 15x - 40 = 0
$$

Group:

$$
(6x^2 - 16x) + (15x - 40) = 2x(3x - 8) + 5(3x - 8) = (2x + 5)(3x - 8)
$$

So:

$$
(2x + 5)(3x - 8) = 0
$$

Set each factor to zero:

- $ 2x + 5 = 0 $ → $ x = -\frac{5}{2} $
- $ 3x - 8 = 0 $ → $ x = \frac{8}{3} $

✔ Solutions: $ x = -\frac{5}{2} $, $ x = \frac{8}{3} $

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6. $ 2x^2 + 13x + 15 = 0 $

Again, use AC method: $ a = 2 $, $ b = 13 $, $ c = 15 $

$ ac = 2 \times 15 = 30 $

Find two numbers that multiply to 30 and add to 13.

Try: $ 10 $ and $ 3 $ → $ 10 \times 3 = 30 $, $ 10 + 3 = 13 $ → Yes!

Split the middle term:

$$
2x^2 + 10x + 3x + 15 = 0
$$

Group:

$$
(2x^2 + 10x) + (3x + 15) = 2x(x + 5) + 3(x + 5) = (2x + 3)(x + 5)
$$

So:

$$
(2x + 3)(x + 5) = 0
$$

Set each factor to zero:

- $ 2x + 3 = 0 $ → $ x = -\frac{3}{2} $
- $ x + 5 = 0 $ → $ x = -5 $

✔ Solutions: $ x = -\frac{3}{2} $, $ x = -5 $

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✔ Final Answers:

| Equation | Solutions |
|--------|-----------|
| $ 2x^2 - 2x - 24 = 0 $ | $ x = 4, -3 $ |
| $ x^2 + 10x + 16 = 0 $ | $ x = -8, -2 $ |
| $ 4x^2 + 15x + 9 = 0 $ | $ x = -\frac{3}{4}, -3 $ |
| $ x^2 - 13x + 42 = 0 $ | $ x = 6, 7 $ |
| $ 6x^2 - x - 40 = 0 $ | $ x = -\frac{5}{2}, \frac{8}{3} $ |
| $ 2x^2 + 13x + 15 = 0 $ | $ x = -\frac{3}{2}, -5 $ |

Let me know if you'd like these checked using the quadratic formula!