How do you solve using the completing the square method $x^{2} - \frac{1}{2} x - \frac{3}{16} = 0$ $x^2 - 1/2 x - 3/16 = 0$ ?

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How do you solve using the completing the square method $x^{2} - \frac{1}{2} x - \frac{3}{16} = 0$ $x^2 - 1/2 x - 3/16 = 0$ ?

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Answer 1 · 2016-09-19T08:50:23

$x = \frac{3}{4} or x = - \frac{1}{4}$ $x = 3/4 " or " x= -1/4$

Explanation:

"Completing the Square".

This involves adding the correct value to a quadratic expression to create a perfect square.

Recall: ${(x - 5)}^{2} = x^{2} - 10 x + 25 \leftarrow {(\frac{- 10}{2})}^{2} = + 25$ $(x-5)^2 = x^2 color(tomato)(-10)xcolor(tomato)(+25)" "larr color(tomato)(((-10)/2)^2 = +25)$

This relationship between $b and c$ $color(tomato)(b and c)$ will always exist.

If the value of $c$ $c$ is not the correct one, add on what you need.
However, if you have an equation add to BOTH sides.

$x^{2} - \frac{1}{2} x - \frac{3}{16} = 0 \leftarrow$ $x^2 - 1/2 x - 3/16 = 0" " larr$ check: is ${(\frac{- 1}{2} \div 2)}^{2} = - \frac{3}{16} ?$ $((-1)/2 div 2)^2 = -3/16?$ No

(the last term MUST be positive, because it is squared)

$x^{2} - \frac{1}{2} x - \frac{3}{16} = 0 \leftarrow$ $x^2 - 1/2 x color(blue)(- 3/16) = 0" "larr$ not what we want, move it to RHS

$x^{2} - \frac{1}{2} x +? = \frac{3}{16} \leftarrow$ $x^2 color(magenta)(- 1/2) x " +? " = color(blue)(3/16) "larr$ add on what you DO want

$[- \frac{1}{2} \to \div 2 \to - \frac{1}{4} \to squared = + \frac{1}{16}]$ $[color(magenta)(-1/2) rarr div 2 rarr-1/4 rarr "squared" =color(red)(+1/16)]$

$x^{2} - \frac{1}{2} x + \frac{1}{16} = \frac{3}{16} + \frac{1}{16} \leftarrow$ $x^2 - 1/2 x color(red)(+1/16) = 3/16 color(red)(+1/16) "larr$ add to BOTH sides

Now write the square of the binomial as ${(x + ?)}^{2}$ $(x +?)^2$

${(x - \frac{1}{4})}^{2} = \frac{4}{16} = \frac{1}{4} \leftarrow - \frac{1}{2} \div 2 = - \frac{1}{4}$ $(xcolor(magenta)(-1/4))^2 = 4/16 = 1/4" "larrcolor(magenta)(-1/2 div 2 = -1/4)$

Now solve the equation - isolate $x$ $x$

${(x - \frac{1}{4})}^{2} = \frac{1}{4}$ $(x-1/4)^2 = 1/4$

$x - \frac{1}{4} = \pm \sqrt{\frac{1}{4}} \leftarrow$ $x-1/4 = +-sqrt(1/4)" "larr$ square root both sides

$x = + \sqrt{\frac{1}{4}} + \frac{1}{4} = \frac{1}{2} + \frac{1}{4} = \frac{3}{4}$ $x = +sqrt(1/4) +1/4 = 1/2+1/4 = 3/4$

Or

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

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