\[(4x^2+6x+4)(4y^2-12y+25)=28.
\]
Find the real value of \[(x,y)\].
Find the real value of (x,y)
Re: Find the real value of (x,y)
"Inspiration is needed in geometry, just as much as in poetry." -- Aleksandr Pushkin
Please install LaTeX fonts in your PC for better looking equations,
learn how to write equations, and don't forget to read Forum Guide and Rules.
Please install LaTeX fonts in your PC for better looking equations,
learn how to write equations, and don't forget to read Forum Guide and Rules.
- Tahmid Hasan
- Posts:665
- Joined:Thu Dec 09, 2010 5:34 pm
- Location:Khulna,Bangladesh.
- Tahmid Hasan
- Posts:665
- Joined:Thu Dec 09, 2010 5:34 pm
- Location:Khulna,Bangladesh.
Re: Find the real value of (x,y)
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.
$$- \frac {1}{2} \leq x \leq \frac{ \sqrt{2} -3}{4}$$
$3 \leq y \leq \frac{4 \sqrt {10} +15}{10}$
but the value of $y$ follows a contradivtion.
বড় ভালবাসি তোমায়,মা
Re: Find the real value of (x,y)
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 $\mathbf{y=\frac{3}{2}}$
$\mathbf{\left\{(x+\frac{3}{4})^2+\frac{7}{16}\right\}.\left\{(y-\frac{3}{2})^2+\frac{16}{4}\right\}=\frac{7}{4}}$
Which is possiable only when $\mathbf{x=-\frac{3}{4}}$ and $\mathbf{y=\frac{3}{2}}$
jagdish