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Two positive numbers $p$ and $q$ have the property that their sum is equal to their product. If their difference is $7$, what is $\frac{1}{\frac{1}{p^2}+\frac{1}{q^2}}$? Your answer will be of the form $\frac{a+b\sqrt{c}}{d}$, where $a$ and $b$ don't both share the same common factor with $d$ and $c$ has no square as a factor. Find $a+b+c+d$.

 Jul 25, 2019
 #1
avatar+19913 
0

Can you post that without the Latex $$$$$$ etc..... hard to interpret (correctly)....

 Jul 25, 2019
 #2
avatar+23862 
+2

Two positive numbers \(p\) and \(q\) have the property that their sum is equal to their product.
If their difference is \(7\), what is \(\dfrac{1}{\dfrac{1}{p^2}+\dfrac{1}{q^2}}\)?


Your answer will be of the form \(\dfrac{a+b\sqrt{c}}{d}\),


where \(a\) and \(b\) don't both share the same common factor with \(d\) and \(c\) has no square as a factor.


Find \(a+b+c+d\).

 

\(\begin{array}{|lrcll|} \hline &\mathbf{p+q} &=& \mathbf{pq} \quad | \quad :pq \\\\ &\dfrac{1}{q} + \dfrac{1}{p} &=& 1 \\ &\left(\dfrac{1}{q} + \dfrac{1}{p}\right)^2 &=& 1^2 \\ & \dfrac{1}{q^2} + \dfrac{1}{p^2} + \dfrac{2}{pq} &=& 1 \\ (1)& \mathbf{ \dfrac{1}{q^2} + \dfrac{1}{p^2}} &=& \mathbf{ 1-\dfrac{2}{pq}} \\ \hline \end{array} \)

 

\(\begin{array}{|lrcll|} \hline &\mathbf{p-q} &=& \mathbf{7} \quad | \quad :pq \\\\ &\dfrac{1}{q} - \dfrac{1}{p} &=& \dfrac{7}{pq} \\ &\left(\dfrac{1}{q} - \dfrac{1}{p}\right)^2 &=& \dfrac{49}{(pq)^2} \\ & \dfrac{1}{q^2} + \dfrac{1}{p^2} - \dfrac{2}{pq} &=& \dfrac{49}{(pq)^2} \\ (2)& \mathbf{ \dfrac{1}{q^2} + \dfrac{1}{p^2}} &=& \mathbf{ \dfrac{49}{(pq)^2}+\dfrac{2}{pq}} \\ \hline \end{array} \)

 

\(\begin{array}{|lrcll|} \hline (1)=(2): & \dfrac{1}{q^2} + \dfrac{1}{p^2} = 1-\dfrac{2}{pq} &=& \dfrac{49}{(pq)^2}+\dfrac{2}{pq} \\\\ & 1-\dfrac{2}{pq} &=& \dfrac{49}{(pq)^2}+\dfrac{2}{pq} \\\\ & 1-\dfrac{4}{pq} &=& \dfrac{49}{(pq)^2} \quad | \quad \cdot (pq)^2 \\\\ & (pq)^2-4(pq) &=& 49 \\\\ & (pq)^2-4(pq) -49 &=& 0 \\\\ & (pq) &=& \dfrac{4 \pm \sqrt{16-4\cdot (-49) } } {2} \\ & (pq) &=& \dfrac{4 \pm \sqrt{212} } {2} \\ & (pq) &=& \dfrac{4 \pm \sqrt{53} } {2} \\ & (pq) &=&2 \pm \sqrt{53} \\\\ (3) & \mathbf{(pq)_1} &=& \mathbf{2 + \sqrt{53}} \\ (4) & \mathbf{(pq)_2} &=& \mathbf{2 - \sqrt{53}} \\ \hline \end{array}\)

 

\(\mathbf{(pq)_1 = 2 + \sqrt{53}}\):

\(\begin{array}{|rcll|} \hline \mathbf{ \dfrac{1}{p^2} + \dfrac{1}{q^2}} &=& \mathbf{ 1-\dfrac{2}{(pq)_1}} \\\\ \dfrac{1}{p^2} + \dfrac{1}{q^2} &=& \dfrac{(pq)_1-2}{(pq)_1} \\\\ x=\dfrac{1}{\dfrac{1}{p^2} + \dfrac{1}{q^2}} &=& \dfrac{(pq)_1}{(pq)_1-2} \\\\ x &=& \dfrac{2 + \sqrt{53}}{2 + \sqrt{53}-2} \\\\ x &=& \left( \dfrac{2 + \sqrt{53}}{ \sqrt{53} } \right)\cdot \dfrac{\sqrt{53}}{\sqrt{53}} \\\\ \mathbf{x} &=& \mathbf{\dfrac{53+2\sqrt{53}}{ 53 }} \quad | \quad \dfrac{a+b\sqrt{c}}{d} \\ && a=53,\ b=2,\ c=53,\ d=53 \\ && \mathbf{a+b+c+d} = 53+2+53+53 &=& \mathbf{ 161 } \\ \hline \end{array} \)

 

\(\mathbf{(pq)_2 = 2 - \sqrt{53}}\):

\(\begin{array}{|rcll|} \hline \mathbf{ \dfrac{1}{p^2} + \dfrac{1}{q^2}} &=& \mathbf{ 1-\dfrac{2}{(pq)_2}} \\\\ \dfrac{1}{p^2} + \dfrac{1}{q^2} &=& \dfrac{(pq)_2-2}{(pq)_2} \\\\ x=\dfrac{1}{\dfrac{1}{p^2} + \dfrac{1}{q^2}} &=& \dfrac{(pq)_2}{(pq)_2-2} \\\\ x &=& \dfrac{2 - \sqrt{53}}{2 - \sqrt{53}-2} \\\\ x &=& \dfrac{2 - \sqrt{53}}{ - \sqrt{53} } \\\\ x &=& \left( \dfrac{-2 + \sqrt{53}}{ \sqrt{53} } \right)\cdot \dfrac{\sqrt{53}}{\sqrt{53}} \\\\ \mathbf{x} &=& \mathbf{\dfrac{53-2\sqrt{53}}{ 53 }} \quad | \quad \dfrac{a+b\sqrt{c}}{d} \\ && a=53,\ b=-2,\ c=53,\ d=53 \\ && \mathbf{a+b+c+d} = 53-2+53+53 &=& \mathbf{ 157 } \\ \hline \end{array}\)

 

laugh

 Jul 26, 2019

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