diota2004 wrote:
Is \(a^4+b^4\) less than 32?
(1) \(a + b = 4\)
(2) \(a^2 + b^2 = 8\)
(1) a + b = 4
We can try using some plugging in of numbers. First thing that comes to mind is 'what if a & b are each equal?'. So when we put a=b=2, we get a^4 + b^4 = 16+16=32.
Now if we give different values to them, eg, put a=3, b=1, we have a^4 + b^4 = 81+1 =82, or if we put a=0, b=4, we have a^4 + b^4 = 0+256 =256. Even if we give negative value to one of a or b, it wont matter because powers are even, it will convert to positive only. eg, if a=-1, b=5, we have a^4 + b^4 =1+625=626
So this plugging of numbers tells me that no matter what values to a&b I give, if a+b = 4, then a^4 + b^4 will always be greater than or equal to 32 (least value 32 will occur when a & b are equal). so a^4 + b^4 can never be less than 32. Sufficient
(2) a^2 + b^2 = 8
Note that a^4 = (a^2)^2 and b^4 = (b^2)^2. So we can proceed with same plugging in of numbers. First lets put a^2=b^2 = 4. In this case a^4 + b^4 = 4^2 + 4^2 = 16+16=32. Now if we substitute unequal values of a^2/b^2, eg, a^2 = 7, b^2 = 1, we get a^4 + b^4 = 49+1=50. Again, no matter which values we give to a^2 and b^2, as long as a^2+b^2 = 8, we will always have a^4 + b^4 as greater than or equal to 32 (least value 32 will occur when a^2 & b^2 are equal). So a^4 + b^4 can never be less than 32. Sufficient.
Hence
D answer