Solving a 'Harvard' University entrance exam question

Added: 2026-06-03

[00:00:00]Hello Friends find the value of 'a' and 'b' If a^2-b=133 and b^2-a=133 where 'a' doesn't equal to 'b' let's have a solution say these are first and second equations first step is subtract first equation from second It will be a^2-b-(b^2-a)=133-133 a^2-b-b^2+a=0 rearrange as a^2-b^2+a-b=0 apply formula (a+b)(a-b)+(a-b)=0 (a-b) is same, can be common (a-b)(a+b+1)=0 either a-b=0 or a+b+1=0 where a=b which is not possible because in question, 'a' doesn't equal to 'b' so, this case is rejected a+b+1=0 can be written as a+b=-1 say this is third equation in the next step, I'm going to add first and third equations a^2-b+a+b=133+(-1) here, 'b' cancels a^2+a=133-1 a^2+a=132 a^2+a-132=0 which is a quadratic equation which can be solved by factorization method 12x(-11)=-132 and 12+(-11)=+1 it can be written as a^2+12a-11a-132=0 a(a+12)-11(a+12)=0 (a+12)(a-11)=0 either a+12=0 or a-11=0 a=-12 which is the value of 'a' a=11 which is also the value of 'a' now, we want to find the value of 'b' recall third equation a+b=-1 putting values of 'a' -12+b=-1 and 11+b=-1 b=-1+12 we get the value of 'b=11' and b=-1-11 b=-12 which is also the value of 'b' (a,b)=(-12,11) and (11,-12) in the next step, I'm going to verify a^2-b=133 and also b^2-a=133 putting values of 'a' and 'b' (-12)^2-11=133 and (11)^2-(-12)=133 simplify this 144-11=133 133=133 121+12=133 133=133 you can see here L.H.S=R.H.S which shows that values of 'a' and 'b' satisfies these equations thanks for watching this video please subscribe this channel to get the notification of my new videos ok bye