Question

A2+b=29,a+b2=21 ,what is the value of a

Answer 1

a²+b=29 and a+b²=21.

Also, (a²+b)-(a+b²)=29-21=8,

a²-b²-(a-b)=8,

(a-b)(a+b)-(a-b)=8,

(a-b)(a+b-1)=8, and 8=1×8=2×4.

If a+b-1=8, so a+b=9 and a-b=1, 2a=10, so a=5 (b=4); a²+b=25+4=29✔️ a+b²=5+16=21✔️

If a+b-1=4, so a+b=5 and a-b=2, 2a=7, so a=7/2 (b=3/2); a²+b=49/4+3/2=55/4≠29❌ a+b²=7/2+9/4=23/4≠21❌

If a+b-1=-1, so b=-a and a-b=-8, 2a=-8 and a=-4 (b=4); a²+b=20≠29❌ a+b²=-4+16=12≠21❌

If a+b-1=-1, so b=-a and a-b=-8, 2a=-8 and a=-4 (b=4); a²+b=20≠29❌ a+b²=-4+16=12≠21❌

None of the other integer combinations work.

So one answer appears to be a=5. However, there may be more solutions, not necessarily integers.

Now let's try a different method:

b=29-a²; a+(29-a²)²=21, a+841-58a²+a⁴=21,

a⁴-58a²+a+820=0. We know one solution is a=5 so we can divide by this solution:

5 | 1 0 -58      1    820

     1 5  25 -165 | -820

     1 5 -33 -164 |    0   =a³+5a²-33a-164, which has 3 irrational solutions: 

a=-5.8465, -4.8900, 5.7365 (approx).

a=5 is probably the expected solution.