Solving quadratic equations by completing the square - worksheet with solutions.

Explanation:
We are solving quadratic equations by completing the square. This method turns a quadratic into a perfect square trinomial, which we can then solve easily.

The general steps are:
1. Make sure the equation is in the form $x^2 + bx = c$ (i.e., coefficient of $x^2$ is 1; if not, divide everything by that coefficient).
2. Take half of $b$, square it, and add it to both sides.
3. Write the left side as $(x + \frac{b}{2})^2$.
4. Solve by taking square roots.

Let’s go through each problem one by one.

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1) $x^2 + 4x = 0$

- Half of 4 is 2; square it → $2^2 = 4$
- Add 4 to both sides:
$x^2 + 4x + 4 = 4$
$(x + 2)^2 = 4$
- Take square roots:
$x + 2 = \pm 2$
- So:
$x = -2 + 2 = 0$
or $x = -2 - 2 = -4$

✔ Solutions: $x = 0$, $x = -4$

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2) $x^2 + 8x = 0$

- Half of 8 is 4; square → $16$
- Add 16:
$x^2 + 8x + 16 = 16$
$(x + 4)^2 = 16$
- Square root:
$x + 4 = \pm 4$
- So:
$x = -4 + 4 = 0$
or $x = -4 - 4 = -8$

✔ Solutions: $x = 0$, $x = -8$

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3) $x^2 - 10x = 0$

- Half of $-10$ is $-5$; square → $25$
- Add 25:
$x^2 - 10x + 25 = 25$
$(x - 5)^2 = 25$
- Square root:
$x - 5 = \pm 5$
- So:
$x = 5 + 5 = 10$
or $x = 5 - 5 = 0$

✔ Solutions: $x = 0$, $x = 10$

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4) $x^2 + 3x = 0$

- Half of 3 is $1.5$; square → $2.25 = \frac{9}{4}$
- Add $\frac{9}{4}$:
$x^2 + 3x + \frac{9}{4} = \frac{9}{4}$
$(x + \frac{3}{2})^2 = \frac{9}{4}$
- Square root:
$x + \frac{3}{2} = \pm \frac{3}{2}$
- So:
$x = -\frac{3}{2} + \frac{3}{2} = 0$
or $x = -\frac{3}{2} - \frac{3}{2} = -3$

✔ Solutions: $x = 0$, $x = -3$

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5) $x^2 - 7x = 0$

- Half of $-7$ is $-\frac{7}{2}$; square → $\frac{49}{4}$
- Add $\frac{49}{4}$:
$x^2 - 7x + \frac{49}{4} = \frac{49}{4}$
$(x - \frac{7}{2})^2 = \frac{49}{4}$
- Square root:
$x - \frac{7}{2} = \pm \frac{7}{2}$
- So:
$x = \frac{7}{2} + \frac{7}{2} = 7$
or $x = \frac{7}{2} - \frac{7}{2} = 0$

✔ Solutions: $x = 0$, $x = 7$

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6) $x^2 + 6x - 1 = 0$

First, move constant to right:
$x^2 + 6x = 1$

- Half of 6 is 3; square → 9
- Add 9:
$x^2 + 6x + 9 = 1 + 9 = 10$
$(x + 3)^2 = 10$
- Square root:
$x + 3 = \pm \sqrt{10}$
- So:
$x = -3 \pm \sqrt{10}$

✔ Solutions: $x = -3 + \sqrt{10}$, $x = -3 - \sqrt{10}$

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7) $x^2 + 14x + 5 = 0$

Move constant:
$x^2 + 14x = -5$

- Half of 14 is 7; square → 49
- Add 49:
$x^2 + 14x + 49 = -5 + 49 = 44$
$(x + 7)^2 = 44$
- Square root:
$x + 7 = \pm \sqrt{44} = \pm 2\sqrt{11}$
- So:
$x = -7 \pm 2\sqrt{11}$

✔ Solutions: $x = -7 + 2\sqrt{11}$, $x = -7 - 2\sqrt{11}$

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8) $x^2 + 11x - 2 = 0$

Move constant:
$x^2 + 11x = 2$

- Half of 11 is $5.5 = \frac{11}{2}$; square → $\frac{121}{4}$
- Add to both sides:
$x^2 + 11x + \frac{121}{4} = 2 + \frac{121}{4} = \frac{8}{4} + \frac{121}{4} = \frac{129}{4}$
- So:
$(x + \frac{11}{2})^2 = \frac{129}{4}$
- Square root:
$x + \frac{11}{2} = \pm \frac{\sqrt{129}}{2}$
- So:
$x = -\frac{11}{2} \pm \frac{\sqrt{129}}{2} = \frac{-11 \pm \sqrt{129}}{2}$

✔ Solutions: $x = \frac{-11 + \sqrt{129}}{2}$, $x = \frac{-11 - \sqrt{129}}{2}$

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10) $x^2 = -20x$

Bring all terms to left:
$x^2 + 20x = 0$

- Half of 20 is 10; square → 100
- Add 100:
$x^2 + 20x + 100 = 100$
$(x + 10)^2 = 100$
- Square root:
$x + 10 = \pm 10$
- So:
$x = -10 + 10 = 0$
or $x = -10 - 10 = -20$

✔ Solutions: $x = 0$, $x = -20$

Note: Problem 9 seems cut off in the image — we’ll skip it since it's incomplete.

Now let’s double-check a few to be safe:

- For #6: $(x+3)^2 = 10$ → expand: $x^2 + 6x + 9 = 10$ → $x^2 + 6x -1 = 0$ ✔
- For #8: $(x + 11/2)^2 = 129/4$ → expand: $x^2 + 11x + 121/4 = 129/4$ → $x^2 + 11x = 8/4 = 2$ → $x^2 + 11x - 2 = 0$ ✔

All correct.

Final Answer:
1) $x = 0,\ -4$
2) $x = 0,\ -8$
3) $x = 0,\ 10$
4) $x = 0,\ -3$
5) $x = 0,\ 7$
6) $x = -3 \pm \sqrt{10}$
7) $x = -7 \pm 2\sqrt{11}$
8) $x = \frac{-11 \pm \sqrt{129}}{2}$
10) $x = 0,\ -20$