A) {3, 4}
B) {3, -3}
C) {3, 4, -3, -4}
D) {-3, -3}
Correct Answer: C
Solution :
\[{{x}^{2}}+{{y}^{2}}=25\] and\[xy=12\] Þ \[{{x}^{2}}+{{\left( \frac{12}{x} \right)}^{2}}=25\,\,\Rightarrow {{x}^{4}}+144-25{{x}^{2}}=0\] Þ \[({{x}^{2}}-16)({{x}^{2}}-9)=0\]Þ \[{{x}^{2}}=16\]and\[{{x}^{2}}=9\] Þ \[x=\pm \,4\]and\[x=\pm 3\].You need to login to perform this action.
You will be redirected in
3 sec