# algebra problem

Printable View

• Jul 6th 2012, 07:51 PM
earthboy
algebra problem
If $\frac{a}{c}=\frac{b}{d}=\frac{3}{4}$
and $\sqrt{a^2+c^2}-\sqrt{b^2+d^2}=15$
find $ac+bd-ad-bc$

please help(Nod)
• Jul 7th 2012, 12:36 AM
Wilmer
Re: algebra problem
108.
a = 9 + 3k, b = 3k, c = 12 + 4k, d = 4k where k is any integer > 0
• Jul 7th 2012, 08:03 AM
Soroban
Re: algebra problem
Hello, earthboy!

Quote:

$\text{Given: }\:\begin{Bmatrix}[1] & \dfrac{a}{c}\,=\,\dfrac{b}{d}\,=\,\dfrac{3}{4} \\ \\ [2] &\sqrt{a^2+c^2}-\sqrt{b^2+d^2}\:=\:15\end{Bmatrix}$

$\text{Find: }\:[3]\;N \:=\: ac+bd-ad-bc$

$\text{From [1]: }\:\begin{Bmatrix}\frac{a}{c} \,=\,\frac{3}{4} &\Rightarrow& c \,=\,\frac{4}{3}a \\ \\[-4mm] \frac{b}{d} \,=\,\frac{3}{4} & \Rightarrow & d \,=\,\frac{4}{3}b \end{Bmatrix}\;\;[4]$

$\text{Substitute [4] into [3]:}$

$N \;=\;a\left(\tfrac{4}{3}a\right) + b\left(\tfrac{4}{3}b\right) - a\left(\tfrac{4}{3}b\right) - b\left(\tfrac{4}{3}a\right) \;=\;\tfrac{4}{3}a^2 + \tfrac{4}{3}b^2 - \tfrac{4}{3}ab - \tfrac{4}{3}ab$

n . $=\;\tfrac{4}{3}a^2 - \tfrac{8}{3}ab + \tfrac{4}{3}b^2 \;=\;\tfrac{4}{3}(a^2 - 2ab + b^2) \;=\;\tfrac{4}{3}(a-b)^2\;\;[5]$

$\text{Substitute [4] into [2]:}$

. . $\sqrt{a^2 + \left(\tfrac{4}{3}a\right)^2} - \sqrt{b^2 + \left(\tfrac{4}{3}b\right)^2} \;=\;15$

. . $\sqrt{a^2 + \tfrac{16}{9}a^2} - \sqrt{b^2 + \tfrac{16}{9}b^2} \;=\;15 \quad\Rightarrow\quad \sqrt{\tfrac{25}{9}a^2} - \sqrt{\tfrac{25}{9}b^2} \;=\;15$

. . $\tfrac{5}{3}a - \tfrac{5}{3}b \;=\;15 \quad\Rightarrow\quad a - b \;=\;9$

$\text{Substitute into [5]: }\:N \;=\;\tfrac{4}{3}(a-b)^2 \;=\;\tfrac{4}{3}(9^2) \;=\;108$