Question

Rewrite each expression by completing the square.

Answer 1

Answer:

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

$(x-2)^2 -\frac{5}{2} -4=(x-2)^2 -\frac{13}{2}$

2) $x^2 -3x +(\frac{3}{2})^2 +\frac{1}{2} - (\frac{3}{2})^2$

$(x-\frac{3}{2})^2 +\frac{1}{2} -\frac{9}{4}=(x-\frac{3}{2})^2 -\frac{7}{4}$

3) $x^2 +\frac{9}{2}x +(\frac{9}{4})^2 -\frac{15}{4} - (\frac{9}{4})^2$

$(x+\frac{9}{4})^2 -\frac{15}{4} -\frac{81}{16}=(x+\frac{9}{4})^2 -\frac{141}{16}$

4) $c^2 -\frac{5}{4}c +(\frac{5}{8})^2 -\frac{139}{200} - (\frac{5}{8})^2$

$(c-\frac{5}{8})^2 -\frac{139}{200} -\frac{25}{64}=(c-\frac{5}{8})^2 -\frac{1737}{1600}$

5) $n^2 +\frac{1}{4}n +(\frac{1}{8})^2 +\frac{5}{8} - (\frac{1}{8})^2$

$(n+\frac{1}{8})^2 +\frac{5}{8} -\frac{1}{64}=(n+\frac{1}{8})^2 +\frac{39}{64}$

Step-by-step explanation:

1. −2x^2 + 8x + 5

For this case we can begin dividing all the terms by -2 and we got:

$x^2 -4x -\frac{5}{2}[/texAnd if we complete the square we got:[tex] x^2 -4x +(\frac{4}{2})^2 -\frac{5}{2} - (\frac{4}{2})^2$

$(x-2)^2 -\frac{5}{2} -4=(x-2)^2 -\frac{13}{2}$

2. 2.5x^2 − 7.5x + 1.25

For this case we can begin dividing all the terms by 2.5 and we got:

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

And if we complete the square we got:

$x^2 -3x +(\frac{3}{2})^2 +\frac{1}{2} - (\frac{3}{2})^2$

$(x-\frac{3}{2})^2 +\frac{1}{2} -\frac{9}{4}=(x-\frac{3}{2})^2 -\frac{7}{4}$

3. 4 / 3x ^2 + 6x − 5

For this case we can begin dividing all the terms by 4/3 and we got:

$x^2 + \frac{9}{2}x - \frac{15}{4}$

And if we complete the square we got:

$x^2 +\frac{9}{2}x +(\frac{9}{4})^2 -\frac{15}{4} - (\frac{9}{4})^2$

$(x+\frac{9}{4})^2 -\frac{15}{4} -\frac{81}{16}=(x+\frac{9}{4})^2 -\frac{141}{16}$

4. 1000c^2 − 1250c + 695

For this case we can begin dividing all the terms by 1000 and we got:

$c^2 - \frac{5}{4}c + \frac{139}{200}$

And if we complete the square we got:

$c^2 -\frac{5}{4}c +(\frac{5}{8})^2 +\frac{139}{200} - (\frac{5}{8})^2$

$(c-\frac{5}{8})^2 -\frac{139}{200} -\frac{25}{64}=(c-\frac{5}{8})^2 -\frac{487}{1600}$

5. 8n^2 + 2n + 5

For this case we can begin dividing all the terms by 8 and we got:

$n^2 + \frac{1}{4}x + \frac{5}{8}$

And if we complete the square we got:

$n^2 +\frac{1}{4}n +(\frac{1}{8})^2 +\frac{5}{8} - (\frac{1}{8})^2$

$(n+\frac{1}{8})^2 +\frac{5}{8} -\frac{1}{64}=(n+\frac{1}{8})^2 +\frac{39}{64}$