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Difference between two numbers is 4, and their product is 17. Then find the sum of their squares?
Read Solution (Total 4)
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- let the nos. be a and b
given, a-b=4 and a*b=17
a^2+b^2=(a-b)^2+2*a*b
putting the values,we get
16+34=50 - 13 years agoHelpfull: Yes(22) No(9)
- Let the no be x & y
x-y=4 & xy=17(given)
x^2+y^2=(x-y)^2+2xy
put the values,so
x^2+y^2= 4^2 + 2*17
x^2 + y^2= 50 - 9 years agoHelpfull: Yes(4) No(1)
- (x-y)^2=x^2+y^2-2*x*y=16
hence x^2+y^2=50 - 9 years agoHelpfull: Yes(3) No(1)
- Say the numbers are x1 and x2
Given x1-x2=4 --(i) and x1*x2=17--(ii)
now x2=x1/1 putting this value in eq(i)we get x1^2-4x1-17=0.--(iii)
Similarly from eq(i) x1=x2+4 putting this value in eq(ii)we get x2^2+4x1-17=0.---(iv)
So by adding the equations (iii) & (iv) we get
sum of their squares as 51 - 2 Months agoHelpfull: Yes(0) No(1)
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