A) 8, 9
B) 6, 7
C) 13, 14
D) 9, 10
Correct Answer: A
Solution :
Let the number be\[x\]and\[(x+1)\]. According to the question. \[{{x}^{2}}+{{(x+1)}^{2}}=145\] \[\Rightarrow {{x}^{2}}+{{x}^{2}}+2x+1=145\] \[\Rightarrow 2{{x}^{2}}+2x-144=0\] \[\Rightarrow {{x}^{2}}+x-72=0\] \[\Rightarrow {{x}^{2}}+9x-8x-72=0\] \[\Rightarrow x(x+9)-8(x+9)=0\] \[\Rightarrow (x-8)(x+9)=0\] \[\Rightarrow x=8\]because \[x\ne -9\] \[\therefore \]Number = 8 and 9 OR \[{{x}^{2}}+{{(x+1)}^{2}}=145\] \[=64+81={{8}^{2}}+{{9}^{2}}\] \[\Rightarrow x=8\]
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