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Mathematics 15 Online
OpenStudy (calculusxy):

MEDAL!!! Completing the square...

OpenStudy (calculusxy):

\[\large -40 = -6n - 4n^2\]

OpenStudy (calculusxy):

My steps were: \[\frac{ -4n^2 - 6n + 40 }{ -4 } = n^2 + 1.5n - 10 \] \[n^2 + 1.5n -10 = 0 \implies n^2 + 1.5n + (1.5/2)^2 = 10 + (15/2)^2\]

OpenStudy (calculusxy):

@Nnesha

Nnesha (nnesha):

looks good :=))

OpenStudy (calculusxy):

Okay so here's the next step: \[n^2 + 1.5n + 56.25 = 56.25 + 10\] I know that I need to find the square roots of both sides. And that's where I get the problem.

Nnesha (nnesha):

ohh well here is a short cut \[\huge\rm x^2 +bx \rightarrow (x+\frac{b}{2})^2 \] here is an example \[x^2+2x \rightarrow (x+\frac{2}{2})^2 \rightarrow (x+1)^2\] make sense ??

Nnesha (nnesha):

x+b/2 would be the factor of quadratic equation

OpenStudy (calculusxy):

What would we do to the "c" term?

Nnesha (nnesha):

let me rewrite it \[\huge\rm x^2+2x +1 \] factor of this quadratic equation would be (x+b/2)(x+b/2) which you can write it as (x+b/2)^2 \[(x+\frac{2}{2})(x+\frac{2}{2})=(x+ \frac{2}{2})^2\]

Nnesha (nnesha):

remember the AC method where we find two numbers when we multiply them we should get product of AC and when we add/subtract them we should get middle term b

Nnesha (nnesha):

we factored it

Nnesha (nnesha):

it's `.5625`

OpenStudy (calculusxy):

don't i need to make \(\rm n^2 + 1.5n + 56.25\) as a perfect square?

Nnesha (nnesha):

\[n^2 + 1.5n + 0.5625= 0.5625 + 10\] (1.5/2)^2 = 0.5625

Nnesha (nnesha):

yes right in other words factor that quadratic equation

OpenStudy (calculusxy):

So I would find the factors of 0.5625? Wouldn't that take some time?

Nnesha (nnesha):

there is a mistake (1.5/2)^2 = 0.5625 not 56.25

OpenStudy (calculusxy):

Yes sorry

Nnesha (nnesha):

that's why i gave u the shortcutt

Nnesha (nnesha):

when we complete the square: factor of the quadratic equation would always be \[(x+\frac{b}{2})^2\]

OpenStudy (calculusxy):

But I don't understand the shortcut. Can we do it with the problem instead of using another example?

Nnesha (nnesha):

sure

Nnesha (nnesha):

\[\color{ReD}{n^2 + 1.5n + 0.5625}= 0.5625 + 10\] factor of this quadratic equation would always be \[\large\rm (n+\frac{b}{2})(n+\frac{b}{2})\] now replace b with its value

Nnesha (nnesha):

\[\rm Ax^2+Bx+C=0\] quadratic equation where coefficient of `x term ` is b

OpenStudy (calculusxy):

\[(n + \frac{ 1.5 }{ 2 })^2\]

OpenStudy (calculusxy):

\[(n + 7.5)^2\]

Nnesha (nnesha):

looks good 1.5 divide by 2 =.75 so \[\huge\rm (n+.75)^2 = 0.5625+10\]

Nnesha (nnesha):

hmm no 1.5 divide by 2 = 0.75

Nnesha (nnesha):

`0.75 is same as .75`

OpenStudy (calculusxy):

\[(n+ 0.75)^2 = 10.5625\] don't i need to find the square root of 10.5625

Nnesha (nnesha):

looks good

Nnesha (nnesha):

yes take square root both sides to cancel out the square

OpenStudy (calculusxy):

\[n + 0.75 = -3.25 \] \[n + 0.75 = 3.25\]

Nnesha (nnesha):

perfecT!

OpenStudy (calculusxy):

I would have to solve for n from both right?

OpenStudy (calculusxy):

okay gimme a moment...

OpenStudy (calculusxy):

The answer would be \(\rm 2.4, -4\)?

OpenStudy (calculusxy):

@Nnesha

Nnesha (nnesha):

hmm i didn't get 2.4

OpenStudy (calculusxy):

Oh I am sorry. I meant to say 2.5

Nnesha (nnesha):

ah then fantastico!

OpenStudy (calculusxy):

thank you!

Nnesha (nnesha):

no problema or problemo ..

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