Q41 / 79 · NEST (NISER) 2026 · Mathematics
A possible solution of the system of equations
$x^2 - 8xy + 16y^2 = 0$
$(\log_{10} x)^2 + 2(\log_{10} x)(\log_{10} y) + (\log_{10} y)^2 = 4$
is
$x^2 - 8xy + 16y^2 = 0$
$(\log_{10} x)^2 + 2(\log_{10} x)(\log_{10} y) + (\log_{10} y)^2 = 4$
is
- Correct answerA$x = \dfrac{1}{5},\ y = \dfrac{1}{20}$
- B$x = 100,\ y = 25$
- C$x = 40,\ y = 10$
- D$x = \dfrac{4}{25},\ y = \dfrac{1}{16}$
Solution
The first equation is $x^2 - 8xy + 16y^2 = 0$
$(x - 4y)^2 = 0$
$x = 4y$
The second equation is $(\log_{10} x)^2 + 2(\log_{10} x)(\log_{10} y) + (\log_{10} y)^2 = 4$
$(\log_{10} x + \log_{10} y)^2 = 4$
$\log_{10} (xy) = \pm 2$
$xy = 10^2 = 100$ or $xy = 10^{-2} = \dfrac{1}{100}$
Substituting $x = 4y$ into $xy = 100$ gives $4y^2 = 100 \Rightarrow y^2 = 25$. Since $y > 0$, $y = 5$ and $x = 20$.
Substituting $x = 4y$ into $xy = \dfrac{1}{100}$ gives $4y^2 = \dfrac{1}{100} \Rightarrow y^2 = \dfrac{1}{400}$. Since $y > 0$, $y = \dfrac{1}{20}$ and $x = \dfrac{1}{5}$.
Therefore, a possible solution is $x = \dfrac{1}{5}, y = \dfrac{1}{20}$.
Answer: $x = \dfrac{1}{5},\ y = \dfrac{1}{20}$
$(x - 4y)^2 = 0$
$x = 4y$
The second equation is $(\log_{10} x)^2 + 2(\log_{10} x)(\log_{10} y) + (\log_{10} y)^2 = 4$
$(\log_{10} x + \log_{10} y)^2 = 4$
$\log_{10} (xy) = \pm 2$
$xy = 10^2 = 100$ or $xy = 10^{-2} = \dfrac{1}{100}$
Substituting $x = 4y$ into $xy = 100$ gives $4y^2 = 100 \Rightarrow y^2 = 25$. Since $y > 0$, $y = 5$ and $x = 20$.
Substituting $x = 4y$ into $xy = \dfrac{1}{100}$ gives $4y^2 = \dfrac{1}{100} \Rightarrow y^2 = \dfrac{1}{400}$. Since $y > 0$, $y = \dfrac{1}{20}$ and $x = \dfrac{1}{5}$.
Therefore, a possible solution is $x = \dfrac{1}{5}, y = \dfrac{1}{20}$.
Answer: $x = \dfrac{1}{5},\ y = \dfrac{1}{20}$
Hence, the correct answer is option: 'A'.