Find the real value of (x,y)

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prodip
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Find the real value of (x,y)

Unread post by prodip » Sun May 22, 2011 11:29 am

\[(4x^2+6x+4)(4y^2-12y+25)=28.
\]

Find the real value of \[(x,y)\].

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Moon
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Re: Find the real value of (x,y)

Unread post by Moon » Sun May 22, 2011 11:51 am

Try squaring the two terms LHS
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Tahmid Hasan
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Re: Find the real value of (x,y)

Unread post by Tahmid Hasan » Sun May 22, 2011 7:55 pm

does a contradiction work here?
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Tahmid Hasan
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Re: Find the real value of (x,y)

Unread post by Tahmid Hasan » Sun May 22, 2011 8:06 pm

i've come up with this solution
$$- \frac {1}{2} \leq x \leq \frac{ \sqrt{2} -3}{4}$$
$3 \leq y \leq \frac{4 \sqrt {10} +15}{10}$
but the value of $y$ follows a contradivtion.
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jagdish
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Re: Find the real value of (x,y)

Unread post by jagdish » Sat May 28, 2011 2:56 pm

Here $\mathbf{(4x^2+6x+4).(4y^2-12y+25)=28}$

$\mathbf{\left\{(x+\frac{3}{4})^2+\frac{7}{16}\right\}.\left\{(y-\frac{3}{2})^2+\frac{16}{4}\right\}=\frac{7}{4}}$

Which is possiable only when $\mathbf{x=-\frac{3}{4}}$ and $\mathbf{y=\frac{3}{2}}$
jagdish

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