Solutions: Challenging Completing the Square

1. Make the coefficient of $x^2$ equal to 1 by dividing the entire equation by 3:

$x^2 - \frac{10}{3}x + 2 = 0$

2. Move the constant term to the right side:

$x^2 - \frac{10}{3}x = -2$

3. Halve the coefficient of $x$ (which is $-\frac{10}{3}$), square it, and add to both sides:

Half of $-\frac{10}{3}$ is $-\frac{10}{6} = -\frac{5}{3}$. Squaring this gives $(-\frac{5}{3})^2 = \frac{25}{9}$.

$x^2 - \frac{10}{3}x + \frac{25}{9} = -2 + \frac{25}{9}$

4. Factor the left side as a perfect square and simplify the right side:

$(x - \frac{5}{3})^2 = -\frac{18}{9} + \frac{25}{9} = \frac{7}{9}$

5. Take the square root of both sides:

$x - \frac{5}{3} = \pm\sqrt{\frac{7}{9}} = \pm\frac{\sqrt{7}}{3}$

6. Solve for $x$:

$x = \frac{5}{3} \pm \frac{\sqrt{7}}{3}$

1. Make the coefficient of $x^2$ equal to 1. Divide by -2 (or multiply by -1 then divide by 2):

$x^2 - \frac{7}{2}x + 2 = 0$

2. Move the constant term:

$x^2 - \frac{7}{2}x = -2$

3. Halve the coefficient of $x$ (which is $-\frac{7}{2}$), square it, and add to both sides:

Half of $-\frac{7}{2}$ is $-\frac{7}{4}$. Squaring this gives $(-\frac{7}{4})^2 = \frac{49}{16}$.

$x^2 - \frac{7}{2}x + \frac{49}{16} = -2 + \frac{49}{16}$

4. Factor and simplify:

$(x - \frac{7}{4})^2 = -\frac{32}{16} + \frac{49}{16} = \frac{17}{16}$

5. Take the square root:

$x - \frac{7}{4} = \pm\sqrt{\frac{17}{16}} = \pm\frac{\sqrt{17}}{4}$

6. Solve for $x$:

$x = \frac{7}{4} \pm \frac{\sqrt{17}}{4}$

1. Move the constant term:

$x^2 + \frac{2}{3}x = \frac{1}{4}$

2. Halve the coefficient of $x$ (which is $\frac{2}{3}$), square it, and add to both sides:

Half of $\frac{2}{3}$ is $\frac{1}{3}$. Squaring this gives $(\frac{1}{3})^2 = \frac{1}{9}$.

$x^2 + \frac{2}{3}x + \frac{1}{9} = \frac{1}{4} + \frac{1}{9}$

3. Factor and simplify (find a common denominator for the right side, which is 36):

$(x + \frac{1}{3})^2 = \frac{9}{36} + \frac{4}{36} = \frac{13}{36}$

4. Take the square root:

$x + \frac{1}{3} = \pm\sqrt{\frac{13}{36}} = \pm\frac{\sqrt{13}}{6}$

5. Solve for $x$:

$x = -\frac{1}{3} \pm \frac{\sqrt{13}}{6}$

To combine the terms, use a common denominator:

$x = -\frac{2}{6} \pm \frac{\sqrt{13}}{6}$

1. Divide by 5:

$x^2 - \frac{11}{5}x + \frac{3}{5} = 0$

2. Move the constant term:

$x^2 - \frac{11}{5}x = -\frac{3}{5}$

3. Halve $-\frac{11}{5}$ to get $-\frac{11}{10}$. Square it to get $\frac{121}{100}$. Add to both sides:

$x^2 - \frac{11}{5}x + \frac{121}{100} = -\frac{3}{5} + \frac{121}{100}$

4. Factor and simplify (common denominator is 100):

$(x - \frac{11}{10})^2 = -\frac{60}{100} + \frac{121}{100} = \frac{61}{100}$

5. Take the square root:

$x - \frac{11}{10} = \pm\sqrt{\frac{61}{100}} = \pm\frac{\sqrt{61}}{10}$

6. Solve for $x$:

$x = \frac{11}{10} \pm \frac{\sqrt{61}}{10}$

1. Divide by $a$ (assuming $a \neq 0$):

$x^2 + \frac{b}{a}x + \frac{c}{a} = 0$

2. Move the constant term:

$x^2 + \frac{b}{a}x = -\frac{c}{a}$

3. Halve $\frac{b}{a}$ to get $\frac{b}{2a}$. Square it to get $(\frac{b}{2a})^2 = \frac{b^2}{4a^2}$. Add to both sides:

$x^2 + \frac{b}{a}x + \frac{b^2}{4a^2} = -\frac{c}{a} + \frac{b^2}{4a^2}$

4. Factor and simplify the right side (common denominator is $4a^2$):

$(x + \frac{b}{2a})^2 = -\frac{4ac}{4a^2} + \frac{b^2}{4a^2} = \frac{b^2 - 4ac}{4a^2}$

5. Take the square root of both sides:

$x + \frac{b}{2a} = \pm\sqrt{\frac{b^2 - 4ac}{4a^2}} = \pm\frac{\sqrt{b^2 - 4ac}}{\sqrt{4a^2}} = \pm\frac{\sqrt{b^2 - 4ac}}{2a}$

6. Solve for $x$:

$x = -\frac{b}{2a} \pm \frac{\sqrt{b^2 - 4ac}}{2a}$

1. Divide by 4:

$x^2 - \frac{9}{4}x + \frac{2}{4} = 0 \implies x^2 - \frac{9}{4}x + \frac{1}{2} = 0$

2. Move the constant term:

$x^2 - \frac{9}{4}x = -\frac{1}{2}$

3. Halve $-\frac{9}{4}$ to get $-\frac{9}{8}$. Square it to get $\frac{81}{64}$. Add to both sides:

$x^2 - \frac{9}{4}x + \frac{81}{64} = -\frac{1}{2} + \frac{81}{64}$

4. Factor and simplify (common denominator is 64):

$(x - \frac{9}{8})^2 = -\frac{32}{64} + \frac{81}{64} = \frac{49}{64}$

5. Take the square root:

$x - \frac{9}{8} = \pm\sqrt{\frac{49}{64}} = \pm\frac{7}{8}$

6. Solve for $x$:

$x = \frac{9}{8} \pm \frac{7}{8}$

This gives two distinct solutions:

$x = \frac{9+7}{8} = \frac{16}{8} = 2$

$x = \frac{9-7}{8} = \frac{2}{8} = \frac{1}{4}$

1. Move the constant term:

$x^2 - \sqrt{5}x = -1$

2. Halve $-\sqrt{5}$ to get $-\frac{\sqrt{5}}{2}$. Square it to get $(-\frac{\sqrt{5}}{2})^2 = \frac{5}{4}$. Add to both sides:

$x^2 - \sqrt{5}x + \frac{5}{4} = -1 + \frac{5}{4}$

3. Factor and simplify:

$(x - \frac{\sqrt{5}}{2})^2 = -\frac{4}{4} + \frac{5}{4} = \frac{1}{4}$

4. Take the square root:

$x - \frac{\sqrt{5}}{2} = \pm\sqrt{\frac{1}{4}} = \pm\frac{1}{2}$

5. Solve for $x$:

$x = \frac{\sqrt{5}}{2} \pm \frac{1}{2}$

1. Expand the left side and rearrange to the standard quadratic form $ax^2+bx+c=0$:

$x^2 - 5x + x - 5 = 1$

$x^2 - 4x - 5 = 1$

$x^2 - 4x - 6 = 0$

2. Move the constant term:

$x^2 - 4x = 6$

3. Halve $-4$ to get $-2$. Square it to get $4$. Add to both sides:

$x^2 - 4x + 4 = 6 + 4$

4. Factor and simplify:

$(x-2)^2 = 10$

5. Take the square root:

$x-2 = \pm\sqrt{10}$

6. Solve for $x$:

$x = 2 \pm \sqrt{10}$

1. Combine the fractions on the left side (common denominator is $(x-1)(x+2)$):

$\frac{(x+2) + (x-1)}{(x-1)(x+2)} = 1$

$\frac{2x+1}{x^2+x-2} = 1$

2. Multiply both sides by the denominator to clear the fraction:

$2x+1 = x^2+x-2$

3. Rearrange into standard quadratic form:

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

$0 = x^2-x-3$

4. Move the constant term:

$x^2 - x = 3$

5. Halve $-1$ to get $-\frac{1}{2}$. Square it to get $\frac{1}{4}$. Add to both sides:

$x^2 - x + \frac{1}{4} = 3 + \frac{1}{4}$

6. Factor and simplify:

$(x - \frac{1}{2})^2 = \frac{12}{4} + \frac{1}{4} = \frac{13}{4}$

7. Take the square root:

$x - \frac{1}{2} = \pm\sqrt{\frac{13}{4}} = \pm\frac{\sqrt{13}}{2}$

8. Solve for $x$:

$x = \frac{1}{2} \pm \frac{\sqrt{13}}{2}$

An equation has exactly one real solution when the completed square form results in $(x+A)^2 = 0$. This means the discriminant ($b^2-4ac$) is zero.

Let's complete the square:

1. Move the constant term:

$x^2 + (k+2)x = -k$

2. Halve the coefficient of $x$ (which is $k+2$) to get $\frac{k+2}{2}$. Square it to get $(\frac{k+2}{2})^2$. Add to both sides:

$x^2 + (k+2)x + (\frac{k+2}{2})^2 = -k + (\frac{k+2}{2})^2$

3. Factor the left side:

$(x + \frac{k+2}{2})^2 = -k + \frac{(k+2)^2}{4}$

For exactly one real solution, the right side must be zero:

$-k + \frac{(k+2)^2}{4} = 0$

Multiply by 4:

$-4k + (k+2)^2 = 0$

$-4k + k^2 + 4k + 4 = 0$

$k^2 + 4 = 0$

This implies $k^2 = -4$. There are no real values of $k$ for which $k^2 = -4$. This means there are no real values of $k$ for which this equation has exactly one real solution if we are strictly completing the square to find $x$.

Alternative approach (using the discriminant, which is equivalent to the RHS of the completed square):

For one real solution, the discriminant $b^2 - 4ac$ must be equal to 0.

In the equation $x^2 + (k+2)x + k = 0$, we have $a=1$, $b=(k+2)$, and $c=k$.

So, $(k+2)^2 - 4(1)(k) = 0$

$k^2 + 4k + 4 - 4k = 0$

$k^2 + 4 = 0$

$k^2 = -4$

As before, there are no real values of $k$ for which $k^2 = -4$. If we allow for complex numbers, $k = \pm 2i$. However, typically in these problems, "real solution" implies "real $k$".