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How do you solve using the completing the square method $4 t^{2} = 8 t - 1$ $4t^2 = 8t - 1$ ?

2 Answers

Mar 14, 2016

See details below to obtain
$XXX t = 1 \pm \frac{\sqrt{3}}{2}$ $color(white)("XXX")t=1+-sqrt(3)/2$

Explanation:

Given
$XXX 4 t^{2} = 8 t - 1$ $color(white)("XXX")4t^2=8t-1$

Get all terms involving the variable on the left side
$XXX 4 t^{2} - 8 t = - 1$ $color(white)("XXX")4t^2-8t=-1$

Divide both sides by $4$ $4$
$XXX t^{2} - 2 t = - \frac{1}{4}$ $color(white)("XXX")t^2-2t= -1/4$

If $t^{2} - 2 t$ $t^2-2t$ are the first two terms of an expanded squared binomial,
then the third term must be $1$ $1$ .
Complete the square by adding $1$ $1$ to both sides:
$XXX t^{2} - 2 t + 1 = \frac{3}{4}$ $color(white)("XXX")t^2-2t+1 = 3/4$

Re-write the left side as a squared binomial
$XXX {(t - 1)}^{2} = \frac{3}{4}$ $color(white)("XXX")(t-1)^2=3/4$

Taking the square roots:
$XXX t - 1 = \pm \frac{\sqrt{3}}{2}$ $color(white)("XXX")t-1=+-sqrt(3)/2$

Add $1$ $1$ to both sides:
$XXX t = 1 - \frac{\sqrt{3}}{2} and t = 1 + \frac{\sqrt{3}}{2}$ $color(white)("XXX")t=1-sqrt(3)/2 and t=1+sqrt(3)/2$

Mar 14, 2016

$t = 1 \pm \frac{\sqrt{3}}{2}$ $t=1+-(sqrt(3))/(2)$

Explanation:

Move all values with $t$ $t$ to the left hand side.

$4 t^{2} - 8 t = - 1$ $4t^2-8t=-1$

Divide everything by the initial coefficient $4$ $4$ to make $t^{2}$ $t^2$ have a coefficient of 1

$t^{2} - 2 t = - \frac{1}{4}$ $t^2-2t=-1/4$ . Note $b = - 2$ $b=-2$

Add the square of half the coefficient $b$ $b$ to both sides

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

Rewrite in perfect square form and simplify the right hand side

${(t - 1)}^{2} = \frac{3}{4}$ $(t-1)^2=3/4$

Take $\pm$ $+-$ the square root of both sides

$t - 1 = \pm \frac{\sqrt{3}}{2}$ $t-1=+-sqrt(3)/2$

Add one to both sides

$t = 1 \pm \frac{\sqrt{3}}{2}$ $t=1+-sqrt(3)/2$

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