Let f(x)=x2+ax+b. If for all non-zero real xf(x+1x)=f(x)+f(1x)and the roots of f(x)=0 are integers, what is the value of a2+b2?

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Alisha Chambers · Accepted Answer

This is a completely routine matter of writing out the given ingredients; you havef(x)=x2+ax+b&amp;nbsp;&amp;nbsp;&amp;nbsp;and&amp;nbsp;&amp;nbsp;&amp;nbsp;f(x+t1x)=f(x)+f(t1x),for all x, so simply write out what&amp;nbsp;f(x+1x)&amp;nbsp;and&amp;nbsp;f(1x)&amp;nbsp;are.We have&amp;nbsp;f(x+1x)=(x+1x)2+a(x+1x)+b&amp;nbsp;=(x2+2+(1x)2)+ax+a1x+b&amp;nbsp;=(x2+ax+b)+((1x)2+a1x+b)−b+2&amp;nbsp;=f(x)+f(1x)−b+2,and so&amp;nbsp;f(x+1x)=f(x)+f(1x)&amp;nbsp;implies that&amp;nbsp;b=2. The roots of&amp;nbsp;f(x)=0&amp;nbsp;are integers, so0=f(x)=x2+ax+2,has integer roots. These must then be divisors of 2, i.e. at least one of 1, 2, -1 and -2 is a root of f(x)=0. These options correspond to four linear equations in a:0=f(1)=12+a×1+2=3+a0=f(2)=22+a×2+2=6+2a0=f(−1)=(−1)2+a×(−1)+2=3−a0=f(−2)=(−2)2+a×(−2)+2=6−2aand so either&amp;nbsp;a=3&amp;nbsp;or&amp;nbsp;a=−3, hence&amp;nbsp;a2+b2=13

Jaslyn Allison · Accepted Answer

Hints:1) Try say x=1 with the condition given, to immediately get b.2) Use the fact that roots are integers and factors of b to find possible values of a.Even though there is more than one possible value for a, you should find a unique value for a2+b2

Let \(\displaystyle{f{{\left({x}\right)}}}={x}^{{2}}+{a}{x}+{b}\). If for all non-zero real

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