1. ## completing square

help........

i have Three questions which i really don't know how to do
pls help me!!
solve the given quadratic equations by completing the square.
$
(1)----v(v + 2) = 15
$

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$
(2)----5T^2 - 10T + 4 = 0
$

$(3)$In machine design. in finding the outside diameter $D_o$ of a hollow shaft, the equation $D_o^2 - DD_o - 0.25D^2 = 0$ is used Solve for $D_o$ if $D = 3.625cm$

2. Originally Posted by danielwu
$
(1)----v(v + 2) = 15
$
$v^2 + 2v = 15$

Now, $x^2 + 2ax + a^2 = (x + a)^2$

Here we have 2 = 2a so a = 1. Thus add $a^2 = 1$ to both sides of the equation:
$v^2 + 2v + 1 = 15 + 1$

$(v + 1)^2 = 16$ Take the square root of both sides.

$v + 1 = \pm 4$

$v = -1 \pm 4 =$ 3 or -5.

(You should always be in the habit of checking your solutions to make sure they work. As it happens, they both do.)

-Dan

3. Originally Posted by danielwu
(2)----5T^2 - 10T + 4 = 0
[/tex]
$5T^2 - 10T = -4$

$T^2 - 2T = -\frac{4}{5}$

(See previous post)
-2 = 2a, so a = -1. Thus add $a^2 = 1$ to both sides.

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

$(T - 1)^2 = \frac{1}{5}$

$T - 1 = \pm \sqrt{\frac{1}{5}} = \pm \frac{\sqrt{5}}{5}$

$T = 1 \pm \frac{\sqrt{5}}{5}$

-Dan

4. Originally Posted by danielwu
$(3)$In machine design. in finding the outside diameter $D_o$ of a hollow shaft, the equation $D_o^2 - DD_o - 0.25D^2 = 0$ is used Solve for $D_o$ if $D = 3.625cm$
Same method. Get the constant term on the RHS of the equation, find out what "a" is, then add $a^2$ to both sides and go from there.

-Dan

5. Originally Posted by danielwu
alright i have solve it.

but here is another question which i also don't know how to do.
can help me.

..[snip]
Please post new unrelated questions in a new thread, it reduces confusion

Thanks

RonL

(I will split this question out into a new thread)